Oh no! Inequality again?

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#1 h Jul 13, 2022, 2:52 AM • 14 Y

Y by nguyentuantai_, timothywang835, dangerousliri, PHSH, Stuart111, megarnie, suh, S.Ragnork1729, Kabir_Basanta, kamatadu, Schur-Schwartz, deplasmanyollari, ItsBesi, cubres

Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$ , there is exactly one $y \in \mathbb{R}^+$ satisfying $xf(y)+yf(x) \leq 2$

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jasperE3

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jasperE3

#2 h Jul 13, 2022, 3:00 AM • 14 Y

Y by nguyentuantai_, timothywang835, strong_boy, yee5487, Donya, Evan_fan, Gato_combinatorio, angelthirty, Lightosv, I.owais, grboy, Sedro, Frank25, MS_asdfgzxcvb

Claim: $f$ is decreasing
Let $a>b$ . By taking $x=a$ , there is some $c$ with $af(c)+cf(a)\le2$ . By taking $x=c$ , we know that $cf(b)+bf(c)>2$ since $a\ne b$ . Then $2<cf(b)+af(c)$ , so by adding this to $af(c)+cf(a)\le2$ , we get $f(a)<f(b)$ .

Now we can claim that $\boxed{f(x)=\frac1x}$ , which is indeed a solution (by AM-GM, we have $xf(y)+yf(x)\ge2\sqrt{xf(x)yf(x)}=2$ with equality iff $x=y$ ). Let $u\in\mathbb R^+$ be arbitrary, proving $f(u)=\frac1u$ suffices.

Suppose $f(u)>\frac1u$ . Then let $2\ge uf(v)+vf(u)$ by taking $x=u$ . We have:
$2\ge uf(v)+vf(u)>uf(v)+\frac vu\ge2\sqrt{vf(v)}$ by AM-GM, so $vf(v)\le1$ . Then $vf(v)+vf(v)\le2$ , so by taking $x=v$ we find that $u=v$ . This yields a contradiction as $f(v)\le\frac1v$ .

Suppose $f(u)<\frac1u$ , and let $d$ be any positive real. Then $uf(u+du)+(u+du)f(u)>2$ by taking $x=u$ and $y=u+du$ , but:
$uf(u+du)+(u+du)f(u)\le uf(u)+(u+du)f(u)<2+d$ since $f$ is decreasing and $f(u)<\frac1u$ . Then $uf(u+du)+(u+du)f(u)\le2$ , contradiction.

The only remaining option is $f(u)=\frac1u$ as desired.

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Whitewizard314

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Whitewizard314

#3 h Jul 13, 2022, 3:02 AM • 2 Y

Y by nguyentuantai_, timothywang835

Let $g:\mathbb{R^+} \to \mathbb{R^+}$ be a function such that $g(x)=f(x)/x$ for positive real $x$ .
Then the given condition is equivalent to : For every $x \in R^+$ there exists exactly one $y \in \mathbb{R^+}$ such that $g(x)+g(y)\leq 2/xy$ .
Let's call such a pair $(x,y)$ a good pair. Let $G$ be the set of all good pairs. Then obviously $(y,x) \in G$ while $(x,z) \notin G$ for $y\neq z$ .

1) $g(x)$ is strictly decreasing.
Let $0<a<b$ be positive real numbers.
Case 1: $(a,b) \notin G$
Suppose $(b,c) \in G$ . $\therefore$ $(a,c) \notin G$ . $\therefore$ $g(a)+g(c)>2/ac$ and $g(b)+g(c) \leq 2/bc$ . By combining these two, we have $g(a)-g(b)\geq{\frac{2}{c}}({\frac{1}{a}}-{\frac{1}{b}})>0$ . $\therefore$ $g(a)>g(b)$ .
Case 2: $(a,b) \in G$
Let $d \in \mathbb{R^+}$ such that $a<d<b$ . $\therefore$ $(a,d),(d,b)\notin G$
By using the result in Case $1$ we have $g(a)>g(d)$ and $g(d)>g(b)$ . $\therefore$ $g(a)>g(b)$
2) If $(x,y)\in G$ then $g(x)+g(y)=2/xy$
Suppose there exists $(x,y) \in G$ which doesn't hold the above. $\therefore$ $g(x)+g(y)<2/xy$ . $\therefore$ $\exists$ $n>0$ so that $g(x)+g(y)<{\frac{2}{x(y+n)}}$ . But $g(x)+g(y)>g(x)+g(y+n)$ . $\therefore$ $g(x)+g(y+n)<{\frac{2}{x(y+n)}}$ which is a contradiction because $(x,y+n) \notin G$ .
3) $g(x)=1/x^2$ and $f(x)=1/x$ for every $x \in \mathbb{R^+}$ .
We've shown that $g(x)+g(y)\geq2/xy\forall x,y\in\mathbb{R^+}$ .
$x=y$ $\Rightarrow$ $g(x)\geq1/x^2$ . Consider any $(a,b) \in G$ . $\therefore$ $2/ab=g(a)+g(b)\geq1/a^2+1/b^2$ $\Rightarrow$ $a=b$ . $\therefore$ By step 2, $g(x)=1/x^2.$ $\therefore$ $f(x)=1/x$ .
It's not difficult to see that $f(x)=1/x$ is indeed a possible solution.
( $x/y+y/x\leq2\iff x=y$ )

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MarkBcc168

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MarkBcc168

#4 h Jul 13, 2022, 3:04 AM • 12 Y

Y by nguyentuantai_, timothywang835, IAmTheHazard, rama1728, geometry6, aidan0626, sabkx, Mango247, megarnie, Randomization, lksb, CyclicISLscelesTrapezoid

This solution probably is (or close to) the weirdest solution I found on any math Olympiad question. In particular, it hinges on the uncountability of $\mathbb R$ . Please help check if it's correct.

The only answer is $f(x) = \tfrac 1x$ , which satisfies the equation due to AM-GM inequality $\tfrac xy + \tfrac yx \geq 2$ . To prove that there are no other solutions, we first substitute $g(x) = xf\left(\tfrac 1x\right)$ , turning the inequality to
$g\left(\frac 1x\right) + g\left(\frac 1y\right) \leq \frac{2}{xy} \iff g(x)+g(y)\leq 2xy.$ Our intermediate goal is to show that $f(x)\geq x^2$ for all $x>0$ . To do so, consider a very small real number $\delta > 0$ (pick later) and partition $\mathbb R^+$ into strips in form $S_n = [n\delta, (n+1)\delta)$ for $n=0,1,2,\dots$ .

Consider a strip $S_a$ where $a>1000$ . Each number $x\in S_a$ (uncountably infinite numbers) is mapped to another real number $y$ such that $g(x)+g(y)\leq 2xy$ . As there are countably many strips, six of them, $a_1,\dots,a_6$ must get mapped $b_1,\dots,b_6$ lying in the same strip $S_b$ . Now, note that

If $a\ne b$ , then just pick the first three $a_1,a_2,a_3$ and $b_1,b_2,b_3$ . We have $a_i\ne b_j$ for all $i,j$ .
If $a=b$ , then these six numbers must pair up to at least $3$ pairs. Therefore, we can pick $a_1,a_2,a_3$ , $b_1,b_2,b_3$ such that $g(a_i) + g(b_i)\leq 2a_ib_i$ and $a_i\neq b_j$ whenever $i\ne j$ .

Now, we contend that $a$ and $b$ must be close. Note that in what follows, all constants in the $O$ are absolute, not depending on $\delta$ .

Claim: $a-b = O(\sqrt a)$ .

Proof: We have the following table to visualize the argument.
$\begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $\min$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq(a\delta)^2$ & + & $\min$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq(a\delta)^2$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}$ Note that since $g(a_i) + g(a_j) \geq 2a_ia_j \geq 2(a\delta)^2$ , we have that $g(a_i) \geq (a\delta)^2$ for all but one $i$ . Similarly, $g(b_i)\geq (b\delta)^2$ for all but one $i$ . Therefore, for some $i$ , $g(a_i)\geq (a\delta)^2$ and $g(b_i)\geq (b\delta)^2$ . This gives
$\delta^2(a^2+b^2) \leq 2\delta^2(a+1)(b+1) \implies a^2+b^2 \leq 2ab+2a+2b+2,$ giving the desired bound. (Explicitly, $|a-b| < 1+\sqrt{4a+3}$ .) $\blacksquare$

Next, we contend that $g(a_1), g(a_2), g(a_3)$ and $g(b_1), g(b_2), g(b_3)$ must be close together. In particular,

Claim: We have $g(a_i) = \delta^2(a^2+O(a))$ . Similarly, $g(b_i) = \delta^2(a^2+O(a))$ .

Proof: We first define some notation. Let
$\begin{align*} m_a &= \min\{g(a_1), g(a_2), g(a_3)\}, \\ M_a &= \max\{g(a_1), g(a_2), g(a_3)\}, \\ m_b &= \min\{g(b_1), g(b_2), g(b_3)\}, \\ M_b &= \max\{g(b_1), g(b_2), g(b_3)\}. \\ \end{align*}$ Then, we prove that $m_a$ and $m_b$ can't be to small. To do so, we construct the table similar to before
$\begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq 2(a\delta)^2-m_a$ & + & $m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq 2(a\delta)^2-m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}$ This time, we have
$2\delta^2(a^2+b^2) - m_a - m_b\leq 2\delta^2(a+1)(b+1) \implies m_a+m_b\geq 2\delta^2(a^2+O(a)).$ Then, we note that
$M_a + m_b \leq g(a_i) + g(b_i) \leq 2\delta^2(a+1)(b+1) = 2\delta^2(a^2+O(a)),$ and analogously, $m_a+M_b \leq 2\delta^2(a^2+O(a))$ . Therefore, we have three inequalities
$\begin{align*} m_a + m_b &\geq 2\delta^2(a^2+O(a)) \\ m_a + M_b &\leq 2\delta^2(a^2+O(a)) \\ m_b + M_a &\leq 2\delta^2(a^2+O(a)). \end{align*}$ Subtracting the first and second gives $M_b - m_b = \delta^2O(a)$ . However, we also have $M_b + m_b \geq 2(b\delta)^2 = 2\delta^2(a^2+O(a))$ , so $m_b \geq 2\delta^2(a^2+O(a))$ . Analogously, $m_a \geq 2\delta^2(a^2+O(a))$ . Therefore, using the second and third inequalities gives the upper bound $M_a,M_b\leq 2\delta^2(a^2+O(a))$ . $\blacksquare$

Now, we link this to the entire strip $S_a$ .

Claim: $g(x) \geq \delta^2(a^2+O(a))$ for all $x\in S_a$ .

Proof: If $x\in\{a_1,a_2,a_3,b_1,b_2,b_3\}$ , we are already done. Otherwise,
$g(x) + g(a_1) \geq 2xa_1 \geq 2\delta^2a^2,$ giving the conclusion as well. $\blacksquare$

By taking $\delta \to 0$ and $a = \left\lfloor\tfrac x\delta\right\rfloor$ , we have
$g(x) > \delta^2\left(\frac{x^2}{\delta^2} + O\left(\frac{x}{\delta}\right)\right) = x^2 + xO(\delta),$ yielding $g(x) \geq x^2$ for all $x$ . For each $x$ , there exists $y$ such that
$2xy \geq g(x)+g(y)\geq x^2+y^2,$ forcing $x=y$ and $g(x)=x^2$ , done.

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Leartia

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Leartia

#5 h Jul 13, 2022, 3:08 AM • 2 Y

Y by nguyentuantai_, timothywang835

Motivation: You want to have something that lets you perturb $y$ for a given $x$ and use uniqueness to get $f(x)=\dfrac{1}{x}.$
Continuity would be great but this is supposed to be high school level so you naturally try to show monotonicity as the next best thing.
Let $x_1>x_2$ and $y\in\mathbb{R}^+$ be the unique number such that $x_1f(y)+yf(x_1)\leq 2.$
Due to the uniqueness of $y$ we have that $x_2f(y)+yf(x_2)>2.$ Since $2\geq x_1f(y)+yf(x_1)>x_2f(y)+yf(x_1)$ and $x_2f(y)+yf(x_2)>2$ we have $f(x_1)<f(x_2).$ Thus the function is decreasing.
Now perturbation will give you what you want.

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jj_ca888

2726 posts

jj_ca888

#6 h Jul 13, 2022, 3:14 AM • 22 Y

Y by timothywang835, richrow12, FishHeadTail, mathitoo, oolite, CherryMagic, miiirz30, lgkarras, 554183, SerdarBozdag, cj13609517288, aidan0626, sabkx, GustavoPudim, ehuseyinyigit, megarnie, Math_legendno12, poirasss, CyclicISLscelesTrapezoid, EpicBird08, ike.chen, RobertRogo

This could have been a 1/4

. Cute little problem though.

I claim our answer is $\boxed{f(x) = \tfrac 1x}.$

Define $x \sim y$ if $y$ is the unique value that makes the statement true for $x$ . Note that $y \sim x$ too, so this property is commutative.

First, we prove $x \sim x$ for all $x$ . Suppose FTSoC that $x \sim y$ with $x \neq y$ . Then, $x \not \sim x$ and $y \not \sim y$ . Hence $2xf(x) > 2$ and similarly $2yf(y) > 2$ , so $f(x) > \tfrac 1x$ and $f(y) > \tfrac 1y$ . It turns out that $xf(y) + yf(x) > \frac{x}{y} + \frac{y}{x} > 2$ since AM-GM equality can never hold, contradiction, as desired.

So, $x \sim x$ always $\implies f(x) \leq \tfrac{1}{x}$ . Now, suppose some $f(c) < \tfrac 1c$ , so $f(c) = \tfrac {1}{c + \epsilon}$ for some $\epsilon$ . Note that $c \sim c + \epsilon$ then, since $cf(c + \epsilon) + (c+\epsilon)f(c) = cf(c + \epsilon) + 1 \leq 1 + \frac{c}{c + \epsilon} \leq 2$ which is bad since $c$ can only $\sim$ one number, already established to be itself. Hence, $f(x) = \tfrac 1x$ for all $x$ , as desired. $\square$

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hyay

181 posts

hyay

#7 h Jul 13, 2022, 3:19 AM • 2 Y

Y by timothywang835, Nartku

The only solution is $f(x) = \frac{1}{x}$ for all $x$ . AM-GM verifies that $y=x$ is the unique value satisfying $xf(y) + yf(x) \leq 2$ for any $x$ .

First, we show that $xf(x) \leq 1$ for all $x$ . Assume otherwise, then consider $x,y \in \mathbb{R}^+$ with $xf(x) > 1$ satisfying the given inequality. By AM-GM,
$2 \geq xf(y) + yf(x) \geq 2\sqrt{(xf(x))(yf(y))} \implies yf(y) < 1$ Hence, $z=y$ is the unique solution to the inequality
$yf(z) + zf(y) \leq 2$ But we know $z=x$ is also a solution, so $x=y$ , and $xf(x) = yf(y) < 1$ , contradiction.

This implies
$xf(y) + yf(x) > 2 \text{ for all } x \neq y$ Now assume that for some $x$ , there is $c < 1$ such that $f(x) = \frac{c}{x}$ . Let $y=x+\epsilon$ , then
$2 < xf(y) + yf(x) \leq \frac{x}{x+\epsilon} + \frac{c(x+\epsilon)}{x}$ for all $\epsilon\neq 0$ . But $\lim_{\epsilon\to 0} RHS = 1 + c < 2$ , contradiction.

Hence, we are done.

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IndoMathXdZ

691 posts

IndoMathXdZ

#8 h Jul 13, 2022, 3:22 AM • 2 Y

Y by timothywang835, I.owais

2022 IMO/2 wrote:

Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$ , there is exactly one $y \in \mathbb{R}^+$ satisfying
$xf(y) + yf(x) \le 2$

The answer is only $\boxed{f(x) = \frac{1}{x}}$ . This works, as for $y = x$ , we have $2xf(x) = 2$ and for $y \not= x$ , by AM-GM, $xf(y) + yf(x) > 2$ . We'll now prove that there are no other solutions.
Claim 01. For all $x \not= y$ , we have
$\frac{2}{xy} < \frac{f(x)}{x} + \frac{f(y)}{y} \le \frac{1}{x^2} + \frac{1}{y^2}$ Proof. Let us suppose that the unique $y \in \mathbb{R}^+$ satisfying $xf(y) + yf(x) \le 2$ is not $x$ . We therefore conclude that $2xf(x) > 2$ and $2yf(y) > 2$ , i.e. $f(x) > \frac{1}{x}$ and $f(y) > \frac{1}{y}$ . However, this implies
$\frac{2}{xy} \ge \frac{f(x)}{x} + \frac{f(y)}{y} > \frac{1}{x^2} + \frac{1}{y^2} \ge \frac{2}{xy}$ which is a contradiction. Therefore, we must have $xf(x) \le 1$ for all $x \in \mathbb{R}^+$ ; and furthermore for all $x \not= y$ , we must have
$\frac{f(x)}{x} + \frac{f(y)}{y} > \frac{2}{xy}$ as claimed.

Now, consider the function $g: \mathbb{R}^+ \to \mathbb{R}_{\ge 0}$ such that $g(x) = x^2 - xf \left( \frac{1}{x} \right) \ge 0$ for all $x \in \mathbb{R}^+$ . Therefore the condition above rewrites to
$g(a) + g(b) = a^2 + b^2 - a f \left( \frac{1}{a} \right) - b f \left( \frac{1}{b} \right) < (a - b)^2$ for any $a \not= b$ . Let us write this as $Q(a,b) : g(a) + g(b) < (a - b)^2$ .
Claim 02. $g \equiv 0$ .
Proof. We have $0 \le g(a) \le g(a) + g(b) < (a - b)^2$ for all $b \not= a$ . Thus,
$0 \le g(a) \le \lim_{b \to a} (a - b)^2 = 0 \implies g(a) = 0$ for any $a \in \mathbb{R}^+$ , as desired.

This forces $f(x) = \frac{1}{x}$ , as desired.

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MathisWow

82 posts

MathisWow

#9 h Jul 13, 2022, 3:23 AM • 5 Y

Y by timothywang835, JG666, sa2001, Mango247, ehuseyinyigit

This problem is a joke.

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asdf334

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asdf334

#10 h Jul 13, 2022, 3:25 AM • 3 Y

Y by timothywang835, sabkx, Mango247

I am struggling more on the IMO 1/4s than the IMO 2/5s.

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khina

993 posts

khina

#12 h Jul 13, 2022, 3:32 AM • 1 Y

Y by timothywang835

solution?

This problem is quite nice and creative. A good p2!

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BlazingMuddy

281 posts

BlazingMuddy

#15 h Jul 13, 2022, 3:36 AM • 2 Y

Y by timothywang835, teomihai

I think this suits P1 more than P2... I'm disappointed that the uniqueness resolves that quickly for a P2. It's probably nicer as a P1 (compared to 2019 A1?), while P1 could fit as an easy-medium P2, I think.

Write $R^+ = \mathbb{R}^+$ . For each $x \in R^+$ , we claim that the unique $y \in R^+$ satisfying $xf(y) + yf(x) \leq 2$ is $y = x$ . In particular, this gives us $x f(x) \leq 1$ and $x f(y) + y f(x) > 2$ for each $x, y \in R^+$ .

To prove the claim, fix some $x \in R^+$ , and suppose that the unique $y$ is distinct from $x$ . Then, we have $2 xf(x) \leq 2 \to xf(x) \leq 1$ , and similarly $y f(y) \leq 1$ . But then $2xy < xy (x f(y) + y f(x)) = x^2 y f(y) + y^2 x f(x) \leq 1 + 1 = 2$ , a contradiction.

Now, for any $x \in R^+$ , denote $g(x) = 1 - x f(x)$ . We have $g(x) \geq 0$ for each $x \in R^+$ , and also $2xy < xy (x f(y) + y f(x)) = x^2 (1 - g(y)) + y^2 (1 - g(x)) \iff x^2 g(y) + y^2 g(x) < (x - y)^2$ for each $x, y > 0$ . For fixed $y$ , this means $y^2 g(y) < (y + c)^2 g(y) \leq (y + c)^2 g(y) + y^2 g(y + c) < c^2$ for any $c > 0$ . Thus $y^2 g(y) < c$ for any $c > 0$ (for example, by replacing $c$ with $\sqrt{c}$ ). Since $g(y) \geq 0$ and $y > 0$ , this forces $g(y) = 0 \to y f(y) = 1 \to f(y) = \dfrac{1}{y}$ for any $y > 0$ .

P.S. Writing $\mathbb{R}^+$ as $R^+$ is actually deliberate; I think the above solution should work even if we are working over the set of positive elements of a linear ordered ring $R$ . In the case where $R$ is not a field, no such function $f$ exists; otherwise, if $R$ is a field, then $f(x) = 1/x$ is the unique solution. This is why I do not use division at all in the above solution (almost...). An easier solution would start with dividing the original inequality by $xy$ as in #8.

One can still prove $y^2 g(y) < c$ for any $c > 0$ , for example, as follows: if $c \geq 1$ then $y^2 g(y) < 1^2 \leq c$ ; otherwise $y^2 g(y) < c^2 < c$ .

This post has been edited 3 times. Last edited by BlazingMuddy, Jul 13, 2022, 3:41 AM
Reason: Typo again

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MarkBcc168

1593 posts

MarkBcc168

#16 h Jul 13, 2022, 3:36 AM • 3 Y

Y by timothywang835, Euler365, NobleNeon111

I wonder if the following variant breaks some solutions. My countability solution still works if we take more $a_i$ 's and $b_i$ 's.

Quote:

Determine all $f : \mathbb R^+ \to \mathbb R^+$ such that for each $x\in\mathbb R^+$ , there are at least one but at most $2022$ positive real numbers $y$ such that $xf(y)+yf(x)\leq 2$ .

This post has been edited 1 time. Last edited by MarkBcc168, Jul 13, 2022, 3:38 AM

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DNCT1

235 posts

DNCT1

#17 h Jul 13, 2022, 3:52 AM • 2 Y

Y by timothywang835, paul.thai.1803

Hope this is true
Let $y_x>0$ be the unique element for every $x>0$ we have $xf(y)+yf(x)\leq 2$ .
Assume that $y_x\neq x$ then we have $2xf(x)>2\implies f(x)>\frac{1}{x}$ for some $x>0 (1)$ .
Then by $AM-GM$ : $2\ge xf(y_x)+y_xf(x)>xf(y_x)+\frac{y_x}{x}\ge 2\sqrt{y_xf(y_x)}\implies f(y_x)<\frac{1}{y_x}$ By $(1)$ we see that $y_{y_{x}}=y_x$ , otherwise we get $f(y_x)>\frac{1}{y_x}$ , a contracdition.
Let $x\to y_x$ then we have $y_xf(y)+f(y_x)y>2\quad\forall y>0,y\neq y_x$ .
But since $x\neq y_x$ so $y_xf(x)+xf(y_x)>2$ , a contracditon. This inffer that we get $y_x=x\quad\forall x>0$ and so
$xf(y)+yf(x)>2\quad\forall x,y>0 / x\ne y , \ f(x)\leq \frac{1}{x}$ That is $\displaystyle\frac{2-\frac{x}{y}}{y}\leq \frac{2-xf(y)}{y}<f(x)\le \frac{1}{x}\quad\forall x,y>0, x\neq y\ (2)$ $\frac{2-\frac{y}{x}}{x}-\frac{1}{x}<f(y)-f(x)<\frac{1}{y}-\frac{2-\frac{x}{y}}{y}\quad\forall x,y>0,x\neq y$ Thus $\lim_{y\to x} f(y)=f(x)\quad\forall x>0$ so $f$ is continous on $\mathbb{R^+}$ .
By $(2)$ and squeeze theorem we have $f(y)=\lim_{x\to y}f(x)=\frac{1}{y}\quad\forall y>0$ So $f(x)=\frac{1}{x}\quad\forall x>0$ and we're done.

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rama1728

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rama1728

#18 h Jul 13, 2022, 3:54 AM • 2 Y

Y by timothywang835, NobleNeon111

I didn't expect this.....

Let $g(x)$ be the unique positive real satisfying the given inequality.

I claim that $f$ is decreasing. Assume that $f(a)\ge f(b)$ and $a\ge b$ for some two positive reals, then note that $g(a)f(b)+bf(g(a))\le g(a)f(a)+af(g(a))\le 2$ contradicting uniqueness of $g(a)$ . Therefore, $f$ is strictly decreasing.

Now, assume that there exists a positive real $x_0$ for which $f(x_0)<\frac{1}{x_0}$ . Then, clearly $g(x_0)=x_0$ (since it satisfies the inequality and must be unique). That means $(x_0+\varepsilon)f(x_0)+x_0f(x_0+\varepsilon)>2$ for $\varepsilon>0$ . Now, since $f(x_0)<\frac{1}{x_0}$ by assumption, we must have $\frac{\varepsilon}{x_0}+x_0f(x_0+\varepsilon)>1$ or $f(x_0+\varepsilon)>\frac{x_0-\varepsilon}{x_0^2}$ And as $f$ is decreasing: $f(x_0)>f(x_0+\varepsilon)>\frac{x_0-\varepsilon}{x_0^2}$ so $x_0^2f(x_0)>x_0-\varepsilon$ which is a contradiction, as we tend $\varepsilon\rightarrow 0^+$ , because $x_0^2f(x_0)<x_0$ .

This implies that $f(x)\ge \frac{1}{x}$ for all $x$ . Assume now that there exists a positive real $t$ for which $f(t)>\frac{1}{t}$ . Then, $2\ge g(t)f(t)+tf(g(t))>\frac{g(t)}{t}+\frac{t}{g(t)}\ge 2$ a contradiction. Therefore, we must have $f(x)=\frac{1}{x}$ for all $x$ .

Note that this solution works, since we have the unique positive real satisfying $\frac{x}{g(x)}+\frac{g(x)}{x}\le 2$ is $g(x)=x$ .

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Sammy27

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Sammy27

#107 h Apr 19, 2024, 9:11 PM • 1 Y

Y by Eka01

Solution

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ihatemath123

3430 posts

ihatemath123

#108 h Apr 29, 2024, 5:25 AM

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Let the $y$ -value of a positive real $x$ be the unique $y$ for which $xf(y) + yf(x) \leq 2$ .

Claim: the $y$ -value of each $x$ is just $x$ .

Proof: Suppose there exists some $a$ which has a $y$ -value of $b \neq a$ . So, $af(a) + af(a) > 2 \implies f(a) > \tfrac{1}{a}$ . So, $2 \sqrt{\left( a f(b) \right) \cdot \left( \frac{b}{a} \right)} \leq a f(b) + \frac{b}{a} < a f(b) + bf(a) \leq 2 \implies f(b) < \frac{1}{b},$ which is a contradiction since $b$ has two possible $y$ -values, namely itself and $a$ .

As a consequence of this claim, $f(x) \leq \tfrac{1}{x}$ for all $x$ .

Claim: $f(x) = \tfrac{1}{x}$ .

Proof: Suppose there exists some $a$ for which $f(a) < \tfrac{1}{a}$ . Then, we have
$2 > a + f(a) \geq \tfrac{a}{f(a)} \cdot f(a) + a \cdot f \left( \frac{a}{f(a)} \right),$ so $\tfrac{a}{f(a)} = a \implies f(a) = 1$ . Thus, $a < 1$ . However, now plugging in $\sqrt{a}$ into the initial assertion gives us a problem: since
$\sqrt{a} f(a) + a f ( \sqrt{a} ) = \sqrt{a} + a f ( \sqrt{a} ) \leq \sqrt{a} + \sqrt{a} < 2,$ $a$ is a $y$ -value for $\sqrt{a}$ , contradiction.

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sami1618

873 posts

sami1618

#109 h May 18, 2024, 6:34 PM

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The only solution is $\boxed{f(x)=1/x}$ which can be seen to work.

Claim: The unique value of $y$ is $x$
Assume not that $y\neq x$ satisfied the condition. Then we must have $2<xf(x)+xf(x)$ and $2<yf(y)+yf(y)$ so we can not have $xf(y)+yf(x)\leq 2$ .

Corollary: $f(x)\leq 1/x$

Claim: $f(x)=1/x$
Assume there exists an $a$ such that $f(a)<1/a$ then choosing a $b$ very close to $a$ yields a contradiction. Formal proof sown in above solutions.

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Jndd

1417 posts

Jndd

#110 h Jul 26, 2024, 11:23 PM

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Since there is exactly one $y$ for each $x$ such that $xf(y)+yf(x)\leq 2$ , this means that $xf(y)+yf(x)>2$ for every $y$ but one. Now, suppose that for some $x$ , the value of $y$ such that $xf(y)+yf(x)\leq 2$ is not equal to $x$ , meaning $xf(y)+yf(x)>2$ for $y=x$ . This gives $f(x)>\frac{1}{x}$ . However, this implies $xf(y)+yf(x) > \frac{x}{y}+\frac{y}{x}\geq 2$ by AM-GM, meaning there is no $y$ such that $xf(y)+yf(x)\leq 2$ , contradiction.

Hence, when $y=x$ , we must have $xf(y)+yf(x)\leq 2$ , giving $f(x)\leq \frac{1}{x}$ . Suppose that $f(x)=\frac{1}{x}$ was not true for all $x$ , meaning there exists some $a$ such that $f(a)=\frac{1}{a}-c$ . Then, for any $x=a+\varepsilon$ , with $\varepsilon > 0$ we would have $2<xf(a)+af(x)=x\left(\frac{1}{a}-c\right)+af(x)\leq \frac{x}{a}-xc+\frac{a}{x},$ which gives us $\frac{a+\varepsilon}{a}+\frac{a}{a+\varepsilon}-(a+\varepsilon)c>2,$ which can be simplified to $\frac{\varepsilon^2}{a(a+\varepsilon)^2}>c,$ which we see is clearly false for sufficiently small $\varepsilon$ . Thus, we must have $f(x)=\frac{1}{x}$ for all $x$ , and it is easy to check this satisfies our inequality conditions by AM-GM.

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Markas

105 posts

Markas

#111 h Aug 20, 2024, 8:06 PM

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We will show that $f(x) = \frac{1}{x}$ is the only answer and it is obviously works if x = y. Now let us have f such that each $x \in R^+$ has a friend y with $xf(y) + yf(x) \le 2$ . By symmetry it is obvious that y is also the friend of x.

Claim: Every number is its own friend.

Proof: FTSOC let $a \neq b$ be friends. Then we have that $af(a) + af(a) > 2$ $\Rightarrow$ $f(a) > \frac{1}{a}$ . Similarly, $f(b) > \frac {1}{b}$ . Using this into the starting inequality we get that $2 \ge af(b) + bf(a) > \frac {a}{b} + \frac {b}{a} \ge 2$ (the last one follows by AM-GM), which is impossible. So now we have that $f(x) \le \frac{1}{x}$ for all x and $xf(y) + yf(x) > 2$ for all $x \neq y$ . Let $y = x + \varepsilon$ for $x>0$ and $\varepsilon > 0$ . So now we get that $2 < xf(x + \varepsilon) + (x + \varepsilon)f(x) \le \frac{x}{x + \varepsilon} + (x + \varepsilon)f(x)$ $\Rightarrow$ we get that $f(x) > \frac{x + 2\varepsilon}{(x + \varepsilon)^2} = \frac{1}{x + \frac{\varepsilon^2}{x + 2\varepsilon}}$ . Since we have that for all $\varepsilon > 0$ this gives us $f(x) \ge \frac{1}{x}$ $\Rightarrow$ $\frac{1}{x} \ge f(x) \ge \frac{1}{x}$ $\Rightarrow$ $f(x) = \frac{1}{x}$ is the only solution, which we already checked that works so we are ready.

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lksb

163 posts

lksb

#112 h Dec 6, 2024, 10:19 PM

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MarkBcc168 wrote:

This solution probably is (or close to) the weirdest solution I found on any math Olympiad question. In particular, it hinges on the uncountability of $\mathbb R$ . Please help check if it's correct.

The only answer is $f(x) = \tfrac 1x$ , which satisfies the equation due to AM-GM inequality $\tfrac xy + \tfrac yx \geq 2$ . To prove that there are no other solutions, we first substitute $g(x) = xf\left(\tfrac 1x\right)$ , turning the inequality to
$g\left(\frac 1x\right) + g\left(\frac 1y\right) \leq \frac{2}{xy} \iff g(x)+g(y)\leq 2xy.$ Our intermediate goal is to show that $f(x)\geq x^2$ for all $x>0$ . To do so, consider a very small real number $\delta > 0$ (pick later) and partition $\mathbb R^+$ into strips in form $S_n = [n\delta, (n+1)\delta)$ for $n=0,1,2,\dots$ .

Consider a strip $S_a$ where $a>1000$ . Each number $x\in S_a$ (uncountably infinite numbers) is mapped to another real number $y$ such that $g(x)+g(y)\leq 2xy$ . As there are countably many strips, six of them, $a_1,\dots,a_6$ must get mapped $b_1,\dots,b_6$ lying in the same strip $S_b$ . Now, note that

If $a\ne b$ , then just pick the first three $a_1,a_2,a_3$ and $b_1,b_2,b_3$ . We have $a_i\ne b_j$ for all $i,j$ .
If $a=b$ , then these six numbers must pair up to at least $3$ pairs. Therefore, we can pick $a_1,a_2,a_3$ , $b_1,b_2,b_3$ such that $g(a_i) + g(b_i)\leq 2a_ib_i$ and $a_i\neq b_j$ whenever $i\ne j$ .

Now, we contend that $a$ and $b$ must be close. Note that in what follows, all constants in the $O$ are absolute, not depending on $\delta$ .

Claim: $a-b = O(\sqrt a)$ .

Proof: We have the following table to visualize the argument.
$\begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $\min$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq(a\delta)^2$ & + & $\min$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq(a\delta)^2$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}$ Note that since $g(a_i) + g(a_j) \geq 2a_ia_j \geq 2(a\delta)^2$ , we have that $g(a_i) \geq (a\delta)^2$ for all but one $i$ . Similarly, $g(b_i)\geq (b\delta)^2$ for all but one $i$ . Therefore, for some $i$ , $g(a_i)\geq (a\delta)^2$ and $g(b_i)\geq (b\delta)^2$ . This gives
$\delta^2(a^2+b^2) \leq 2\delta^2(a+1)(b+1) \implies a^2+b^2 \leq 2ab+2a+2b+2,$ giving the desired bound. (Explicitly, $|a-b| < 1+\sqrt{4a+3}$ .) $\blacksquare$

Next, we contend that $g(a_1), g(a_2), g(a_3)$ and $g(b_1), g(b_2), g(b_3)$ must be close together. In particular,

Claim: We have $g(a_i) = \delta^2(a^2+O(a))$ . Similarly, $g(b_i) = \delta^2(a^2+O(a))$ .

Proof: We first define some notation. Let
$\begin{align*} m_a &= \min\{g(a_1), g(a_2), g(a_3)\}, \\ M_a &= \max\{g(a_1), g(a_2), g(a_3)\}, \\ m_b &= \min\{g(b_1), g(b_2), g(b_3)\}, \\ M_b &= \max\{g(b_1), g(b_2), g(b_3)\}. \\ \end{align*}$ Then, we prove that $m_a$ and $m_b$ can't be to small. To do so, we construct the table similar to before
$\begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq 2(a\delta)^2-m_a$ & + & $m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq 2(a\delta)^2-m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}$ This time, we have
$2\delta^2(a^2+b^2) - m_a - m_b\leq 2\delta^2(a+1)(b+1) \implies m_a+m_b\geq 2\delta^2(a^2+O(a)).$ Then, we note that
$M_a + m_b \leq g(a_i) + g(b_i) \leq 2\delta^2(a+1)(b+1) = 2\delta^2(a^2+O(a)),$ and analogously, $m_a+M_b \leq 2\delta^2(a^2+O(a))$ . Therefore, we have three inequalities
$\begin{align*} m_a + m_b &\geq 2\delta^2(a^2+O(a)) \\ m_a + M_b &\leq 2\delta^2(a^2+O(a)) \\ m_b + M_a &\leq 2\delta^2(a^2+O(a)). \end{align*}$ Subtracting the first and second gives $M_b - m_b = \delta^2O(a)$ . However, we also have $M_b + m_b \geq 2(b\delta)^2 = 2\delta^2(a^2+O(a))$ , so $m_b \geq 2\delta^2(a^2+O(a))$ . Analogously, $m_a \geq 2\delta^2(a^2+O(a))$ . Therefore, using the second and third inequalities gives the upper bound $M_a,M_b\leq 2\delta^2(a^2+O(a))$ . $\blacksquare$

Now, we link this to the entire strip $S_a$ .

Claim: $g(x) \geq \delta^2(a^2+O(a))$ for all $x\in S_a$ .

Proof: If $x\in\{a_1,a_2,a_3,b_1,b_2,b_3\}$ , we are already done. Otherwise,
$g(x) + g(a_1) \geq 2xa_1 \geq 2\delta^2a^2,$ giving the conclusion as well. $\blacksquare$

By taking $\delta \to 0$ and $a = \left\lfloor\tfrac x\delta\right\rfloor$ , we have
$g(x) > \delta^2\left(\frac{x^2}{\delta^2} + O\left(\frac{x}{\delta}\right)\right) = x^2 + xO(\delta),$ yielding $g(x) \geq x^2$ for all $x$ . For each $x$ , there exists $y$ such that
$2xy \geq g(x)+g(y)\geq x^2+y^2,$ forcing $x=y$ and $g(x)=x^2$ , done.

best solution i've seen in a long time

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AshAuktober

924 posts

AshAuktober

#113 h Dec 18, 2024, 2:08 PM

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We claim that the only working function is $f(x) = \frac 1x$ . We can see this works because we have $\frac xy + \frac yx \ge 2$ by AM-GM, with equality holding only for $x = y$ .

Motivated by this solution, we try to find the value $y$ for which $xf(y)+yf(x) \leq 2$ holds (call it the friend of $x$ ).
We claim that the only friend of $x$ is $x$ itself (that is, $x$ is a very lonely person, just like me

).
Indeed assume otherwise that $a$ and $b$ are friends, with $a \ne b$ . Note that $f(a) > \frac 1a$ and $f(b) > \frac 1b$ by putting $(a,a)$ and $(b, b)$ . But putting $(a, b)$ , we have
$af(b) + bf(a) > \frac ab + \frac ba \ge 2,$ a contradiction to the fact that $a$ and $b$ are friends. $\square$

Now we effectively have $f(x) \le \frac 1x$ as well as $xf(y) + yf(x) > 2 \forall x \ne y$ .
The second equation gives $f(y) > \frac 2x - \frac{yf(x)}{x} \ge \frac 2x - \frac y{x^2} \forall x \ne y$
But now taking $\lim_{x \to y}$ , we have $f(y) \ge \frac 1y$ . Therefore $f(x) \ge \frac 1x \forall x$ , and using this with the first equation we have $\boxed{f(x) = \frac 1x}$ as desired. $\square$

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redred123

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redred123

#114 h Dec 22, 2024, 5:12 PM

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cubres

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cubres

#115 h Jan 2, 2025, 9:21 PM

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Storage - grinding IMO problems

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blueprimes

303 posts

blueprimes

#117 h Mar 2, 2025, 10:22 PM

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We claim the only solution is $f(x) \equiv \dfrac{1}{x}$ , which works since $\dfrac{x}{y} + \dfrac{y}{x} \le 2 \implies x = y$ from the equality case of AM-GM.

Call an unordered pair $\{ x, y \}$ prime if $x f(y) + y f(x) \le 2$ . By the condition of the problem, all distinct prime pairs are disjoint. If $\{ x, x \}$ is a prime pair, we have $f(x) \le \dfrac{1}{x}$ . On the other hand, if $\{ x, y \}$ is a prime pair where $x \ne y$ , then $f(x) > \dfrac{1}{x}$ . Clearly these are also if and only if statements by dichotomy.

$\textbf{Claim:}$ All prime pairs are of the form $\{ x, x \}$ .
For the sake of contradiction, assume otherwise. Then some prime pair $\{ a, b \}$ exists with $a \ne b$ , implying $f(a) > \dfrac{1}{a}$ . So
$a f(b) + \dfrac{b}{a} < a f(b) + b f(a) \le 2 \le \dfrac{a}{b} + \dfrac{b}{a} \implies f(b) < \dfrac{1}{b}.$ Thus, $\{b, b \}$ is also a prime pair, hence we reach a contradiction.

Now we have $f(x) \le \dfrac{1}{x}$ for all $x$ and $x f(y) + y f(x) > 2$ for $x \ne y$ . If $f(x) \not \equiv \dfrac{1}{x}$ then there exists some $a$ satisfying $a \ne \dfrac{1}{f(a)}$ . Asserting $(x, y) = \left( a, \dfrac{1}{f(a)} \right)$ into our equation we obtain $f \left( \dfrac{1}{f(a)} \right) > \dfrac{1}{a}$ but
$f \left( \dfrac{1}{f(a)} \right) \le f(a) \le \dfrac{1}{a}$ so we reach a contradiction. Thus $f(x) \equiv \dfrac{1}{x}$ as needed.

This post has been edited 1 time. Last edited by blueprimes, Mar 2, 2025, 10:32 PM
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Mathandski

711 posts

Mathandski

#118 h Mar 5, 2025, 8:34 PM

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Solved this during another Socratic Seminar in English class

Subjective Rating (MOHs)
Please contact westskigamer@gmail.com if there is an error with any of my solution for cash bounties by 3/18/2025.

Solution

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ravengsd

6 posts

ravengsd

#119 h Mar 13, 2025, 10:17 AM

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The only solution is $f(x)=\frac{1}{x}, \forall x \in \mathbb{R}^{+}$ , which works. We will prove now this is the only function satisfying the given condition.
Call $y$ the "pair" of $x$ is $y$ is the only number satisfying $xf(y)+yf(x)\leq 2$ .

Claim: $x$ is the pair of $x, \forall x \in \mathbb{R}^{+}$ .
Proof: FTSoC suppose there exists $x\in \mathbb{R}^{+}$ such that its pair is a number $y \neq x$ .
Since $x$ is not the pair of $x$ we have $2xf(x) > 2$ , therefore $xf(x)>1$ . Similarly, $2yf(y)>2$ so $yf(y)>1$ .
Multiplying these gives: $xf(x)\cdot yf(y)>1$ $\Rightarrow \sqrt{xf(y)\cdot yf(x)}>1$ $\Rightarrow xf(y)+yf(x) >2$ The last inequality follows from AM-GM.
However, this is a contradiction, because $x$ and $y$ are a pair. So the claim is proved. $\square$

From the claim above we have:
$xf(x)\leq 1$ and $xf(y)+yf(x) >2, \forall x, y \in \mathbb{R}^{+}, x\neq y$ .
Claim: $f(x)=\frac{1}{x}, \forall x \in \mathbb{R}^{+}$ .
Proof: Suppose FTSoC that there exists some $x$ such that $xf(x)<1$ . Then, $x< \frac{1}{f(x)}$ so, the second relation gives $xf(\frac{1}{f(x)}) > 1$ , therefore:
$f(\frac{1}{f(x)}) > \frac{1}{x} > f(x)$ $\Rightarrow \frac{1}{f(x)} f(\frac{1}{f(x)})>1$ And clearly, this is a contradiction.
So, $\forall x\in \mathbb{R}$ we must have $f(x)=\frac{1}{x}$ , as desired. $\blacksquare$

This post has been edited 2 times. Last edited by ravengsd, Mar 13, 2025, 12:27 PM

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Ilikeminecraft

298 posts

Ilikeminecraft

#120 h Mar 14, 2025, 2:41 PM

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Claim: The single possible value of $y$ as stated in the problem is $x.$
Proof: Assume not. Observe that we require $f(x)>\frac1x.$ If $f(x)\leq \frac1x,$ then we get $xf(x) + xf(x) \leq 2,$ contradiction. However, we get that $xf(y) + yf(x) > \frac xy + \frac yx > 2.$ Contradiction.

Thus, we also get the bound $f(x)\leq \frac1x.$ Now, assume $f(x) = \frac1{x + \varepsilon}$ for some $\varepsilon > 0.$ We get that $(x + \varepsilon)f(x) + xf(x+\varepsilon) \leq 1 + \frac x{x+\varepsilon} <2,$ contradiction. Thus, $f\equiv\frac1x.$

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quantam13

90 posts

quantam13

#121 h Yesterday at 3:53 PM

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The only answer is $\boxed{f(x)\equiv \frac{1}{x}}$ which clearly works by AM-GM.
For every positive real $a$ , denote its friend as the unique real number $b$ such that $af(b)+bf(a)\le 2$ . Note that friendship is mutual.

Claim: Every positive real is its own friend
Proof of claim: FTSOC, say $a\neq b$ are friends. This mean that $a$ is not its own friend, and neither is $b$ , so we get that $2af(a)> 2$ and $2bf(b)>2$ . This means that $af(b)+bf(a)>\frac{a}{b}+\frac{b}{a}\underset{\text{AMGM}}{>}2$ contradicting the fact that $a$ and $b$ are friends.

Now the condition simplifies to the two given statments $f(x) \le \frac1x \text{ for all $x$}\qquad xf(y) + yf(x) > 2 \text{ for all $x \neq y$}.$ In particular, if we take $x>0$ and $\varepsilon > 0$ we have $\begin{align*} 2 &< xf(x+\varepsilon) + (x+\varepsilon)f(x) \le \frac{x}{x+\varepsilon} + (x+\varepsilon) f(x) \\ \implies f(x) &> \frac{x+2\varepsilon}{(x+\varepsilon)^2} = \frac{1}{x + \frac{\varepsilon^2}{x+2\varepsilon}}. \end{align*}$ But this holds for all $\varepsilon > 0$ , so if we take $\varepsilon \rightarrow 0$ , we get $f(x) \ge \frac1x$ as well which finishes.

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Maximilian113

497 posts

Maximilian113

#122 h Yesterday at 4:45 PM

Y by

For each $x,$ let $g(x)$ be that unique number that satisfies the inequality. Then clearly $g(x)$ is a bijection and involution. Now, for the sake of a contradiction assume that there are two sets $A \cup B = \mathbb R^+$ such that every $x \in A$ has $g(x)=x,$ and every $x \notin B$ has $g(x) \neq x,$ and $|A|, |B| > 0.$

Then for any $x \in B,$ $g(x) \notin A \implies g(x) \in B$ by involution. But for these values of $x$ we have $xf(x)+xf(x)>2 \implies f(x) > \frac{1}{x}.$ Hence since the range of $g(x)$ is a subset of $B,$ we have for $x \in B,$ $2 \geq yf(x)+xf(y) > y/x+x/y \geq 2$ by AM-GM, a contradiction. Therefore $|B| = 0,$ so $g(x)=x$ for all $x.$ It follows that $f(x) \leq \frac{1}{x} \, \forall x, \text{ and } \, yf(x)+xf(y)>2 \, \forall x \neq y.$ Hence, for $x \neq y,$ $2<yf(x)+xf(y) \leq \frac{y}{x}+\frac{x}{y}.$ As $y$ approaches $x,$ by AM-GM equality must hold. It follows that $\boxed{f(x)=\frac{1}{x} \, \forall \, x \in \mathbb R^+}.$ This solution clearly works.

This post has been edited 1 time. Last edited by Maximilian113, Yesterday at 4:46 PM
Reason: typo

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