Square Root Equality

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djmathman

7936 posts

djmathman

#1 h May 1, 2013, 10:02 PM • 8 Y

Y by Davi-8191, sioly, megarnie, jhu08, HWenslawski, son7, Adventure10, Mango247

Find all real numbers $x,y,z\geq 1$ satisfying $\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.$

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tenniskidperson3

2376 posts

tenniskidperson3

#2 h May 1, 2013, 10:03 PM • 8 Y

Y by anser, myh2910, mijail, megarnie, HWenslawski, son7, Adventure10, Mango247

My Solution

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thecmd999

2860 posts

thecmd999

#3 h May 1, 2013, 10:05 PM • 5 Y

Y by megarnie, Adventure10, Mango247, Sedro, and 1 other user

DARN FORGOT TO CONSIDER EQUALITY CASE

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aZpElr68Cb51U51qy9OM

1600 posts

aZpElr68Cb51U51qy9OM

#4 h May 1, 2013, 10:06 PM • 2 Y

Y by megarnie, Adventure10

Without loss of generality, let $\min(x, y, z) = z$ . Then we have to find all solutions $(x, y, z)$ , where $x, y, z \ge 1$ such that $\sqrt{z+xyz} = \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.$ Let $x-1=r^2, y-1 = s^2, z-1=t^2$ for $r, s, t \ge 0$ and $r, s, t \in \mathbb{R}$ . Then $\sqrt{(t^2+1)(1+(r^2+1)(s^2+1))} = r+s+t$ or $(rst)^2+(rs)^2+(st)^2+(tr)^2+t^2+2-2(rs+st+tr) = 0$ . By AM-GM,
$\begin{align*}0 &= (rst)^2+(rs)^2+(st)^2+(tr)^2+t^2+2-2(rs+st+tr) \\&= rst^2\left(rs+\frac{1}{rs}\right) +(rs)^2+(st)^2+(tr)^2+2-2(rs+st+tr) \\& \ge (rs)^2+(st)^2+(tr)^2+2rst^2 + 2 - 2(rs+st+tr) \\&= (rs-1)^2+(st+tr)^2+1-2(st+tr) \\&= (rs-1)^2+(st+tr-1)^2 \ge 0\end{align*}$
so equality is reached when $rs = 1$ and $st+tr = 1$ , or $\sqrt{x-1}\sqrt{y-1} = 1$ and $(\sqrt{x-1}+\sqrt{y-1})\sqrt{z-1} = 1$ . Solving for $y$ in the former yields $y = \frac{x}{x-1}$ , and squaring the latter yields
$\begin{align*} ((x-1)+(y-1)+2\sqrt{x-1}\sqrt{y-1})(z-1) = 1 \implies (x+y)(z-1) = 1 \implies z = \frac{1}{x+y} +1 = \frac{x^2+x-1}{x^2} \end{align*}$
Thus the solutions are the permutations of $(x, y, z) = \left(a, \frac{a}{a-1}, \frac{a^2+a-1}{a^2}\right)$ for $a > 1$ .

Would I lose one point if I wrote $a \ge 1$ instead of $a > 1$ ?

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kakkokari

91 posts

kakkokari

#5 h May 1, 2013, 10:07 PM • 2 Y

Y by megarnie, Adventure10

tenniskidperson3 wrote:

My Solution

x cannot be equal to 1

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thecmd999

2860 posts

thecmd999

#6 h May 1, 2013, 10:07 PM • 2 Y

Y by megarnie, Adventure10

That's what I did except I concluded that there were no solutions

(Specifically, I set $s=b+c$ and $p=bc$ .)

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va2010

1276 posts

va2010

#7 h May 1, 2013, 10:08 PM • 8 Y

Y by ingenio, champion999, megarnie, aidan0626, Adventure10, and 3 other users

Yay! I never knew that I was smart enough to get #6 on a USAJMO!

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q12

17 posts

q12

#8 h May 1, 2013, 10:09 PM • 3 Y

Y by megarnie, Adventure10, Mango247

hmm I got $(x, y, z)$ can be any permutation of $(\frac{1}{n}, \frac{1}{1-n}, -n^2+n+1)$ where $0<n<1$
though I might have screwed up entirely

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superpi83

1416 posts

superpi83

#9 h May 1, 2013, 10:10 PM • 9 Y

Y by r31415, steve123456, yugrey, myh2910, megarnie, Adventure10, Mango247, and 2 other users

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=585&t=290260 darn.

I saw this one before LOL.

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r31415

746 posts

r31415

#10 h May 1, 2013, 10:13 PM • 3 Y

Y by megarnie, Adventure10, Mango247

outline

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TahsinSaffat

186 posts

TahsinSaffat

#11 h May 1, 2013, 10:14 PM • 4 Y

Y by sflorin, megarnie, Adventure10, Mango247

Do you lose points if you have the right answer but have c=k^2+1 in the parameterization?

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pi37

2079 posts

pi37

#12 h May 1, 2013, 10:14 PM • 3 Y

Y by megarnie, Adventure10, Mango247

Solution

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colinhy

751 posts

colinhy

#13 h May 1, 2013, 10:16 PM • 3 Y

Y by megarnie, Adventure10, Mango247

I wrote the triples as x, y, z without saying that $\sqrt{y-1}\sqrt{z-1} = 1$ . In the last minute, I also added on for all y, z, since I forgot what my proof was about so I'll probably lose a lot of points on that. If you mention the product identity, though, you should be fine.

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BOGTRO

5818 posts

BOGTRO

#14 h May 1, 2013, 10:18 PM • 4 Y

Y by wwwrqnojcm, megarnie, Adventure10, Alex-131

How very original.

Spent 3.5 hours trying to prove no solutions, then decided to randomly take quadratic formula, and it worked magically yay.

Actual solution

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Dynamite127

450 posts

Dynamite127

#15 h May 1, 2013, 10:28 PM • 3 Y

Y by megarnie, Adventure10, Mango247

Can someone please check my solution?

solution

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bora_olmez

277 posts

bora_olmez

#57 h Aug 21, 2021, 2:06 PM

Y by

Nice and easy, solved with TheProblemIsSolved
Solution

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rafaello

1079 posts

rafaello

#61 h Aug 31, 2021, 1:16 AM • 1 Y

Y by Mango247

Let $x=a+1$ , $y=b+1$ and $z=c+1$ , and wlog $c=\min(a,b,c)$ . Note that $a,b,c\geq 0$ .
Thus, we get the following equality,
$abc+ab+bc+ca+c+2=2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca},$ which is equivalent to $c(a+1)(b+1)-\sqrt{c}(2\sqrt{b}+2\sqrt{c})+2-2\sqrt{ab}+ab=0.$ Solving the equality wrt $\sqrt{c}$ , we get that
$\begin{align*} \sqrt{c}&=\frac{2\sqrt{a}+2\sqrt{b}\pm\sqrt{4(\sqrt{a}+\sqrt{b})^2-4(a+1)(b+1)(2+ab-2\sqrt{ab})}}{2(a+1)(b+1)}\\&=\frac{\sqrt{a}+\sqrt{b}\pm\sqrt{(\sqrt{a}+\sqrt{b})^2-(a+1)(b+1)(2+ab-2\sqrt{ab})}}{(a+1)(b+1)}. \end{align*}$ However, note that $\begin{align*}(a+1)(b+1)(ab+2-2\sqrt{ab})&=(a+1)(b+1)((\sqrt{ab}-1)^2+1)\\&\geq (a+1)(b+1)\\&=a+b+ab+1\geq a+b+2\sqrt{ab}\\&=(\sqrt{a}+\sqrt{b})^2, \end{align*}$ where the equality holds iff $ab=1$ . Indeed the equality must hold as otherwise the discriminant is negative. Thus, note that
$\sqrt{c}=\frac{a+\frac{1}{a}}{(1+a)(1+\frac{1}{a})}\implies c=\frac{a}{(a+1)^2}.$ We conclude that our solutions are in the form $(x,y,z)=(a+1,\frac{a+1}{a},\frac{a^2+3a+1}{(a+1)^2})$ for any $a>0$ .

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megarnie

5552 posts

megarnie

#62 h Nov 24, 2021, 4:02 PM • 1 Y

Y by OronSH

WLOG $x\le y\le z$ . So now $\sqrt{x+xyz}=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$
We also have $\sqrt{x+xyz}=\sqrt{x}\cdot\sqrt{yz+1}$ .

By Cauchy-Schwarz, $((x-1)+1)(1+yz))\ge (\sqrt{x-1}+\sqrt{yz})^2,$ so $\sqrt{x}\cdot\sqrt{yz+1}\ge \sqrt{x-1}+\sqrt{yz}$ since $x,y,z\ge 1$ .

Applying Cauchy-Schwarz again, we have $(y-1)+1)(1+(z-1))\ge(\sqrt{y-1}+\sqrt{z-1})^2,$ so $\sqrt{yz}\ge \sqrt{y-1}+\sqrt{z-1}$ .

So we have $\sqrt{x+xyz}\ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1},$ which implies equality must hold.

Thus, $\frac{\sqrt{x-1}}{1}=\frac{1}{\sqrt{yz}}\text{ and }\frac{\sqrt{y-1}}{1}=\frac{1}{\sqrt{z-1}}$

These simplify to $x-1=\frac{1}{yz}$ and $y-1=\frac{1}{z-1}$ . This also implies that $z-1=\frac{1}{y-1}$ .

Now, we set $y=c$ , to get $z=\frac{1}{c-1}+1=\frac{c}{c-1}$ and $x=\frac{c-1}{c^2}+1=\frac{c^2+c-1}{c^2}$ .

So our final answer is $\boxed{\left(\frac{c^2+c-1}{c^2},c,\frac{c}{c-1}\right)}$ and its permutations for some $c>1$ .

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IAmTheHazard

5000 posts

IAmTheHazard

#63 h Mar 4, 2022, 4:44 PM

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Let $x=a^2+1,y=b^2+1,z=c^2+1$ and WLOG let $\min\{x,y,z\}=x$ . Then the equation becomes
$\begin{align*} \sqrt{(a^2+1)((b^2+1)(c^2+1)+1)}&=a+b+c\\ (a^2+1)(b^2c^2+b^2+c^2+2)&=a^2+b^2+c^2+2ab+2bc+2ca\\ (b^2c^2+b^2+c^2+1)a^2-2(b+c)a+(b^2c^2-2bc+2)&=0\\ (b^2+1)(c^2+1)a^2-2(b+c)a+(1+(bc-1)^2)&=0. \end{align*}$ By the quadratic formula (discriminant) this has a real solution in $a$ only if
$4(b+c)^2 \geq 4(b^2+1)(c^2+1)(1+(bc-1)^2) \implies (b+c)^2 \geq (b^2+1)(c^2+1) \implies 0 \geq (bc-1)^2,$ which is only true for $bc=1$ , where the discriminant is zero. Then letting $b=k$ for $k \geq 0$ , we have $c=\tfrac{1}{k}$ and
$a=\frac{2(b+c)}{(b^2+1)(c^2+1)}=\frac{k+\frac{1}{k}}{k^2+2+\frac{1}{k^2}}=\frac{1}{k+\frac{1}{k}},$ which is always the least among $b,c$ , so we in fact have $\min\{x,y,z\}=x$ . From here we extract
$(x,y,z)=\left(1+\frac{1}{(k+\tfrac{1}{k})^2},1+k^2,1+\frac{1}{k^2}\right)~\forall k \geq 0$ and permutations, which can be checked to work. $\blacksquare$

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DottedCaculator

7326 posts

DottedCaculator

#64 h Apr 4, 2022, 6:19 PM

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Claim:
We have $\sqrt{x-1}+\sqrt{y-1}\leq\sqrt{xy}$ with equality if and only if $x+y=xy$ .

Proof:
Squaring, it suffices to show that $x+y-2+2\sqrt{(x-1)(y-1)}\leq xy$ , so if $a=xy-x-y$ , then $2\sqrt{a+1}\leq a+2$ , which follows from AM-GM since $\frac{(a+1)+1}2\geq\sqrt{a+1}$ , with equality if and only if $a=0$ , or $x+y=xy$ .

Therefore, if we assume without loss of generality $x\geq y\geq z$ , then we have
$\begin{align*} \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}&\leq\sqrt{(xy+1)-1}+\sqrt{z-1}\\ &\leq\sqrt{xyz+z}\\ &=\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz}). \end{align*}$ The only time both sides can be equal is when $x+y=xy$ and $(xy+1)z=xy+z+1$ . This is equivalent to $(x-1)(y-1)=1$ and $z=1+\frac1{xy}$ . Therefore, we have $x=1+m$ , $y=1+\frac1m$ , and $z=1+\frac m{(m+1)^2}$ , so the solutions are $\boxed{\left(1+m,1+\frac1m,1+\frac m{(m+1)^2}\right)}$ and permutations.

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ZETA_in_olympiad

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ZETA_in_olympiad

#65 h Jul 4, 2022, 6:54 AM

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By CS note that for any $r,s,t:$ $(1+(r+s)^2)(1+t^2)\geq (r+s+t)^2$ Set $(x,y,z)=(1+r^2,1+s^2,1+t^2)$ so $(r+s)^2\geq (1+r^2)+(1+s^2)\iff (rs-1)^2\leq 0$ Therefore $rs=1$ and $t=\tfrac{1}{r+s}$ Our solution is $\{x,y,z\}=\left \{1+r^2,1+\frac{1}{r^2}, 1+\left(\frac{r}{r^2+1}\right )^2\right \}$ which works.

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Alcumusgrinder07

95 posts

Alcumusgrinder07

#66 h Aug 5, 2022, 11:46 PM

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STORAGE Click to reveal hidden text

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brainfertilzer

1831 posts

brainfertilzer

#67 h May 3, 2023, 8:10 PM • 1 Y

Y by megarnie

Let $(x,y,z) = (a^2 + 1, b^2 + 1, c^2+1)$ with $a,b,c\ge 0$ and WLOG $a = \min(a,b,c)$ . Then, we have
$\sqrt{a^2 + 1+ (a^2 + 1)(b^2+ 1)(c^2+1)} = a + b+ c$ $\implies (a^2 + 1)((b^2 + 1)(c^2 + 1) + 1) = (a +b + c)^2.$ Note that $(b^2 + 1)(c^2 + 1) = ((bc)^2 + 1) + b^2 + c^2\ge (2bc) + b^2 + c^2 = (b+ c)^2$ , with equality when $bc = 1$ . Thus, we have
$(a + b +c)^2\ge (a^2 + 1)(1 + (b+c)^2).$ However, by Cauchy, we have $(a^2 + 1)(1 + (b+ c)^2)\ge (a + b + c)^2$ , so equality in both the above inequality as well as the inequality $(b^2 + 1)(c^2 +1)\ge (b + c)^2$ must hold. That is, $(a, 1)$ and $(1, b+c)$ are proportional and $bc = 1$ . This is easy to solve, and we get
$(x,y,z) = \left(1 + \left(\frac{1}{b + \frac{1}{b}}\right)^2, 1 + b^2, 1 + \frac{1}{b^2}\right),$ as well as permutations, for any $b\in\mathbb{R}_{>0}$ .

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shendrew7

793 posts

shendrew7

#68 h May 26, 2024, 7:48 PM

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Our claimed solution set, which we can confirm to work, as the permutations of
$\boxed{\left(\frac{y^2+y-1}{y^2}, y, \frac{y}{y-1}\right), \quad y > 1}.$
Now assume WLOG $x = \min(x,y,z)$ . Our equation is then
$\begin{align*} \sqrt{x+xyz} &= \sqrt{x-1} + (\sqrt{y-1} + \sqrt{z-1}) \\ &\leq \sqrt{x-1} + \sqrt{(1+(y-1)) \cdot ((z-1)+1)} \\ &= \sqrt{x-1} + \sqrt{(yz+1)-1} \\ &\leq \sqrt{(1+(x-1)) \cdot ((yz+1)+1)} \\ &= \sqrt{x+xyz}, \end{align*}$
where we have the inequalities by Cauchy. We must have the two equality cases
$(x-1)(y-1) = (x-1)(yz+1) = 1,$
from which we get the solutions above. $\blacksquare$

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pie854

243 posts

pie854

#69 h Aug 24, 2024, 5:30 PM

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Nice and easy!

Let $(x,y,z)=\left (a^2+1,b^2+1,c^2+1\right )$ and suppose wlog that $0\leq a\leq b\leq c$ . Then $\sqrt{(a^2+1)+(a^2+1)(b^2+1)(c^2+1)}=a+b+c \ \Rightarrow \ a^2\left (b^2c^2+b^2+c^2+1\right )-a(2b+2c)+\left (b^2c^2-2bc+2\right )=0.$ The discriminant is $4(b+c)^2-4\left (b^2c^2-2bc+2\right )\left (b^2c^2+b^2+c^2+1\right )=-4(bc-1)^2\left(c^2b^2+c^2+b^2+2\right )\leq 0$ and so it follows that $c=1/b$ . And we also find $a=\frac{b}{b^2+1}$ .

Conversely it's easy to check that $(x,y,z)=\left(\frac{t^2}{(t^2+1)^2}+1,t^2+1,\frac 1{t^2}+1\right)$ works.

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cursed_tangent1434

565 posts

cursed_tangent1434

#70 h Dec 7, 2024, 4:19 PM

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Pretty cool and conceptual kind of problem. We claim that the answer is $(x,y,z)=\left(\frac{a}{(a+1)^2}+1,a+1,\frac{1}{a}+1\right)$ for all positive real numbers $a$ . It is (not) easy to see that these triples so. Now we shall show that they are the only ones. For this, we first show the following inequality.

Claim : For all positive real numbers $a$ and $b$ ,
$\sqrt{a} + \sqrt{b} \le \sqrt{(a+1)(b+1)}$ with equality if and only if $ab=1$ .

Proof : Note that,
$\begin{align*} (\sqrt{ab}-1)^2 & \ge 0\\ ab -2\sqrt{ab} +1 & \ge 0\\ a + 2 \sqrt{ab} + b & \le ab + a + b + 1\\ (\sqrt{a}+\sqrt{b})^2 & \le (a+1)(b+1)\\ \sqrt{a} + \sqrt{b} & \le \sqrt{(a+1)(b+1)} \end{align*}$ as we set out to show, with equality clearly if and only if $ab=1$ .

Now, say the minimum among $x,y,z$ is $x$ . Then applying the above lemma twice we note,
$\sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} \ge \sqrt{x-1} + \sqrt{yz} \ge \sqrt{x(yz+1)}=\sqrt{x+xyz}$ with equality if and only if equality holds in each inequality. Thus, we are left to solve the system of equations,
$(y-1)(z-1)=1 \text{ and } (x-1)yz=1$ First, note that the first equation rewrites to,
$yz=y+z$ Let $y=a+1$ and $z=b+1$ for non-negative reals $a$ and $b$ . Then,
$ab+a+b+1=(a+1)(b+1)=(a+1)+(b+1)=a+b+2$ which simplifies to $ab=1$ . Thus, we have that
$y=a+1 \text{ and } z= \frac{1}{a}+1$ Substituting these results into the second equation we obtain that,
$x=\frac{a}{(a+1)^2}+1$ and indeed all solutions take the desired forms.

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EpicBird08

1741 posts

EpicBird08

#71 h Jan 3, 2025, 2:57 AM

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Video solution

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HamstPan38825

8857 posts

HamstPan38825

#72 h Mar 12, 2025, 3:39 AM • 1 Y

Y by anduran

what the heck is this

Let $x \leq y \leq z$ . By Cauchy-Schwarz,
$\begin{align*} \left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2 &\leq (z+yz-z+1)\left(\frac{z-1}z + \frac 1z + x - 1\right) \\ &= xyz+x. \end{align*}$ Equality occurs when $\tfrac{z-1}{z^2} = \tfrac 1{z^2(y-1)} = x-1$ , which yields the solution set $(x, y, z) = \left(\frac{z^2+z-1}{z^2}, \frac z{z-1}, z\right)$ for all $z \geq 1$ . One can check that this does satisfy the above equation and $x \leq y \leq z$ , as needed.

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