p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21

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p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21

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Source: IMO Shortlist 2005 problem A3

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who

#1 h Jul 8, 2006, 4:51 PM • 4 Y

Y by Davi-8191, Adventure10, Mango247, and 1 other user

Four real numbers $p$ , $q$ , $r$ , $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$ . Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2$ .

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#2 h Jul 8, 2006, 7:59 PM • 12 Y

Y by tenplusten, B.J.W.T, rashah76, Vieta827, Adventure10, megarnie, SerdarBozdag, Jalil_Huseynov, teomihai, Quidditch, Mango247, and 1 other user

WLOG assume that $p\geq q\geq r\geq s$ .
Easily $pq+rs+pr+qs+ps+qr\leq 30$ and from the first $pq+rs\geq 10$ . Now put $p+q=t$

Then $t^{2}+(t-9)^{2}\geq 41$ so $(t-4)(t-5)\geq 0$ and since
$t\geq r+s$ we find that $25\leq 21-r^{2}-s^{2}+2pq\leq 21+2(pq-rs)$ and QED

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campos

#3 h Dec 31, 2006, 7:59 PM • 6 Y

Y by dangerousliri, rashah76, Adventure10, Mango247, bin_sherlo, and 1 other user

use that $2(x^{2}+y^{2})=(x+y)^{2}+(x-y)^{2}$ . then,

$84=4(p^{2}+q^{2}+r^{2}+s^{2})=(p+q+r+s)^{2}+(p+q-r-s)^{2}+(p+r-q-s)^{2}+(p+s-q-r)^{2}$

this implies that $(p+q-r-s)^{2}+(p+r-q-s)^{2}+(p+s-q-r)^{2}=3$ , then

$\max\{|p+q-r-s|,|p+r-q-s|,|p+s-q-r|\}\geq 1$ .

suppose wlog that $|p+q-r-s|\geq 1$ , then this implies that

$2(p+q)-9=p+q-r-s\geq 1$ or $2(r+s)-9=r+s-p-q\geq 1$

the first case implies that $p+q\geq 5$ , while the second that $r+s\geq 5$ . suppose wlog that $p+q\geq 5$

then, $21+2(pq-rs)=(p+q)^{2}+(r-s)^{2}\geq 25$ , from where we conclude that $pq-rs\geq 2$ .

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#4 h May 30, 2010, 8:17 PM • 2 Y

Y by Adventure10, Mango247

who wrote:

Four real numbers $p$ , $q$ , $r$ , $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$ . Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2$ .

Let $a \ge b \ge c \ge d$ .
$ab-cd \ge \frac{1}{2}(c+d)(a+b-c-d) \iff$
$2ab+c^2+d^2 \ge ac+ad+bc+bd$ . Since $c^2+d^2 \ge 2cd$ it suffices to prove:
$2ab+2cd \ge ac+ad+bc+bd \iff$
$(a-d)(b-c) + (a-c)(b-d) \ge 0$ .
And:
$3(a+b-c-d)^2 \ge 3 = 4(a^2+b^2+c^2+d^2)-(a+b+c+d)^2 \iff$
$2ab+2cd \ge ac+ad+bc+bd$ is true, so $a+b-c-d \ge 1$ .
Let $a+b-c-d = t$ . Then $\frac{1}{2}(c+d)(a+b-c-d) = \frac{1}{4}(2c+2d)(a+b-c-d) = \frac{1}{4}(9-t)t$ , which is increasing on $(-\infty;4.5]$ so $\frac{1}{4}(9-t)t \ge \frac{1}{4}(9-1)\cdot1 = 2$ , so $ab-cd \ge 2$ .
(There is equality only for $(a,b,c,d) = (3,2,2,2)$ )

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Zhero

#5 h Jul 22, 2010, 8:17 PM • 1 Y

Y by Adventure10

Here's another solution. It's not nearly as elegant, but in my opinion it is much more straightforward:

WLOG, let $p \geq q \geq r \geq s$ . I claim that $pq - rs \geq 2$ . I also claim that the result true not only when $p^2 + q^2 + r^2 + s^2 = 21$ , but also when $p^2 + q^2 + r^2 + s^2 \geq 21$ , provided that $s > \frac{1}{4}$ .

But we may always suppose that $s > \frac{1}{4}$ ! If $s \leq \frac{1}{4}$ , then $p+q+r = 9-s \geq \frac{71}{4}$ . It follows that $p^2 + q^2 + r^2 \geq \left(\frac{p+q+r}{3}\right)^2 \geq \left(\frac{71}{12}\right)^2 > 5^2 = 25 > 21$ , which is a contradiction, so we may indeed always suppose that $s \geq \frac{1}{4}$ .

Replacing $p$ and $q$ with $p_0 = p+q - r$ and $q_0 = r$ preserves $p_0 \geq q_0 \geq r \geq s$ , $p_0 + q_0 + r + s \geq 2$ , and $p_0^2 + q_0^2 + r^2 + s^2 \geq 21$ . In addition, $p_0q_0 - rs = pr + qr - r^2 - rs \leq pq - rs$ (since $(p-r)(q-r) \geq 0$ .) Hence, we may suppose without loss of generality that $q=r$ .

We reformulate our problem as follows: when $p \geq q \geq s \geq \frac{1}{4}$ , $p+2q+s = 9$ , and $p^2 + 2q^2 + s^2 \geq 21$ , then $q(p-s) \geq 2$ .

Let $k = \frac{q-s}{2}$ . Let $p' = p + k$ , $s' = s + k$ , and $q' = q - k$ . We see that $p' + 2q' + s' = 9$ , $q'(p-s') = (q-k)(p-s) \leq q(p-s)$ , and that
$\begin{align*} p'^2 + 2q'^2 + s'^2 &= p^2 + 2q^2 + s^2 + 4k^2 + 2k(p+s-2q) \\ &= p^2 + 2q^2 + s^2 + (p-s)^2 + (p-s)(p+s-2q) \\ &= p^2 + 2q^2 + s^2 + 2(p-q)(q-s) \\ &\geq p^2 + 2q^2 + s^2 \geq 21. \end{align*}$
Observing that $s' = q'$ and that $s' \geq s$ , we see that it is sufficient to prove this inequality when $q=s$ .

We reformulate our problem again: when $p \geq q \geq \frac{1}{4}$ , $p+3q = 9$ , and $p^2 + 3q^2 \geq 21$ , then $pq - q^2 \geq 2$ . $9 = p + 3q \leq 4q$ , so $q \leq \frac{9}{4}$ . $p^2 + 3q^2 = (9-3q)^2 + 3q^2 \geq 21$ , so $12q^2 - 54q + 81 \geq 21$ , so $2q^2 - 9q + 10 \geq 0$ , so $(q-2)(2q-5) \geq 0$ , so $q \leq 2$ or $q \geq \frac{5}{2}$ . But $q \leq \frac{9}{4}$ , so we must have that $q \leq 2$ .

$pq - q^2 = (9-3q)q - q^2 = 9q - 4q^2$ . Since $9q - 4q^2$ is concave in $q$ , it attains its minimum when $q$ is either maximized or minimized. $\frac{1}{4} < q \leq 2$ ; but at both $q = \frac{1}{4}$ and $q = 2$ , $9q - 4q^2 = 2$ , so we see that $pq - q^2$ must always be greater than or equal to 2, which completes our proof.

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#6 h Dec 25, 2010, 2:53 AM • 2 Y

Y by Adventure10, Mango247

I'm posting this for the same reason as Zhero.

Let $p\ge q\ge r\ge s$ as everyone else did, and note that by Cauchy-Schwarz, we have
$0\le 3(21-s^2)-(9-s)^2=2(2s-3)(3-s),$ whence $3/2\le s\le 3$ . In particular, $s>0$ . Now define nonnegative reals $m,n$ such that
$(p+q,p^2+q^2,r+s,r^2+s^2)=(9/2+m,21/2+n,9/2-m,21/2-n).$ Then
$2pq-2rs=(p+q)^2-(r+s)^2-(p^2+q^2)+(r^2+s^2)=18m-2n$ and
$\begin{align*} 0\le(p-q)^2=2(p^2+q^2)-(p+q)^2=(21+2n)-(9/2+m)^2=3/4+2n-9m-m^2\\ 0\le(r-s)^2=2(r^2+s^2)-(r+s)^2=(21-2n)-(9/2-m)^2=3/4-2n+9m-m^2. \end{align*}$ By the latter, we have
$pq-rs-2=9m-n-2\ge9m-(3/8+9m/2-m^2/2)-2=(2m-1)(2m+19)/8.$ Thus it suffices to show that $m\ge1/2$ . But this is clear: since $q\ge r$ , we obtain
$q=(9/2+m)-\sqrt{(p-q)^2}\ge(9/2-m)+\sqrt{(r-s)^2}=r,$ or
$2m\ge\sqrt{(p-q)^2}+\sqrt{(r-s)^2}\ge\sqrt{(p-q)^2+(r-s)^2}=\sqrt{3/2-2m^2},$ whence
$4m^2\ge3/2-2m^2\implies m\ge1/2,$ as desired.

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#7 h Sep 6, 2014, 2:21 AM • 5 Y

Y by Swag00, ValidName, ehuseyinyigit, Adventure10, Mango247

Assume WLOG that $p \ge q \ge r \ge s$ . Note that $pq + pr + ps + qr + qs + rs = \frac{(p + q + r + s)^2 - (p^2 + q^2 + r^2 + s^2)}{2} = 30$ . By the rearrangement inequality, we have that $pq + rs = \max(pq + rs, pr + qs, ps + qr)$ and so $pq + rs \ge 10$ . Letting $x = p + q$ we have that $x^2 + (x - 9)^2 = 21 + 2(pq + rs) \ge 41$ which becomes $(x - 4)(x - 5) \ge 0 \Longrightarrow x \ge 5$ . This means that $2\sqrt{rs} \le r + s \le 4 \Longrightarrow rs \le 4$ . But since $pq + rs \ge 10$ this means that $pq \ge 6$ and so we are done.

Now I want to discuss motivation. After some playing around we see the only equality case is $(p, q, r, s) = (3, 2, 2, 2)$ . Now, looking at this, we want the extreme values of the sum of the two biggest or the two smallest to be greater than $5$ and less than $4$ respectively, so letting $x = p + q$ we want to get exactly the equation $(x - 4)(x - 5) \ge 0$ . Now we want an $x - 9$ in there somewhere, and rearranging it turns out we want $x^2 + (x - 9)^2 \ge 41.$ But this is the same as $pq + rs \ge 10$ , and when looking at a sum like this, the rearrangement inequality should come to mind immediately.

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#8 h Nov 29, 2019, 4:45 AM • 1 Y

Y by Adventure10

Solved with eisirrational. We believe that this solution, although ``ugly" (compared to, say, post 2), is straightforward. (Of course, here, the solution is presented backwards.)

Without loss of generality assume $p \leq q \leq r \leq s$ . The central claim is $p+q \leq 4$ . Set $(x,y) = (p+q,p^2+q^2)$ . Then
$\begin{align*} q &= \frac{1}{2}((p+q) + |p-q|) = \frac{1}{2}\left(x + \sqrt{2y-x^2}\right), \\ r &= \frac{1}{2}((r+s) - |r-s|) = \frac{1}{2}\left((9-x) - \sqrt{2(21-y) - (9-x)^2}\right). \end{align*}$ Then we have $q \leq r$ , which yields
$x + \sqrt{2y-x^2} \leq (9-x) - \sqrt{2(21-y) - (9-x)^2} \implies \sqrt{(2y-x^2)(2(21-y) - (9-x)^2)} \leq 3(x-4)(x-5),$ so $3(x-4)(x-5) \geq 0$ , id est $x \leq 4$ (as $p+q \leq \tfrac{1}{2}(p+q+r+s) = 4.5$ ).

Then by Cauchy-Schwarz,
$(p-4.5)^2 + (q-4.5)^2 \geq \frac{(p+q-9)^2}{2} \geq 12.5.$ But this expands to $p^2 + q^2 - 9p - 9q \geq -28$ , which yields
$\begin{align*} 2 &\leq p^2 + q^2 - 9p - 9q + 30 \\ &= \frac{1}{2}((9-p-q)^2 - (21 - p^2 - q^2)) - pq \\ &= \frac{1}{2}((r+s)^2 - (r^2+s^2)) - pq \\ &= rs - pq, \end{align*}$ as desired. Equality holds when $p+q = 4$ , id est when $(p,q,r,s) = (2,2,2,3)$ .

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#9 h Nov 29, 2019, 5:30 AM • 2 Y

Y by Adventure10, Mango247

nice solutions!

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#10 h Dec 28, 2019, 4:37 AM • 2 Y

Y by teomihai, Adventure10

WLOG $p\ge q\ge r\ge s$ . If $s$ is negative, then $p+q+r\ge 9\implies p+q\ge 6$ . So, $(p+q)^2+(r-s)^2>36>25\implies 2(pq-rs)>4$ Hence, we can assume that $s\ge 0$ . Now, the permutation which maximizes $ab-cd$ is $(p,q,r,s)$ , so assume the contrary, namely that $pq-rs<2$ . Now, consider replacing $(r,s)\to\left(\frac{r+s}{2},\frac{r+s}{2}\right)$ , and $(p,q)$ with suitable numbers such that both conditions are still preserved. Obviously, we still have $p\ge q\ge r\ge s\ge 0$ , and this decreases $pq-rs$ , hence we only need to prove the case where $r=s$ .

In this case, $p+q=9-2s$ , $p^2+q^2=21-2s^2\implies 2pq=6s^2-36s+60\implies (p-q)^2=-8s^2+36s-39$ . Solving, $q=\frac{1}{2}\left(9-2s-\sqrt{-8s^2+36s-39}\right)$ . On the other hand, $pq<2+s^2\implies 3s^2-18s+30<2+s^2\implies s^2-9s+14<0\implies s>2$ . As $s$ is the smallest, it lies in the interval $2<s<2.25$ . As $s$ increases in this range, both $2s$ and $-8s^2+36s-39$ increase, so $q$ decreases. Hence, $q<\frac{1}{2}\left(9-4-\sqrt{1}\right)=2$ , and we get $q<s$ , which is a contradiction, as desired.

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#11 h Apr 4, 2020, 6:52 AM • 3 Y

Y by teomihai, amar_04, A-Thought-Of-God

who wrote:

Four real numbers $p$ , $q$ , $r$ , $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$ . Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2$ .

Assume that $\max\{pq,pr,ps,rq,rs,qs\}=pq,$ and assume on the contrary that $pq<rs+2.$ Since $pq+pr+ps+rq+rs+qs=60,$ hence $pq \geqslant 10.$ In particular, this means $2+rs>10 \implies rs>8>0.$ Then since $p^2,q^2,r^2,s^2$ are positive reals, hence by AM-GM
$\begin{align*} 8^2 < (rs)^2 \leqslant pqrs \leqslant \left( \frac{p^2+q^2+r^2+s^2}{4}\right)^2 \implies 8^2 \cdot 4^2 < 21^2 \end{align*}$ which is clearly false. $\square$

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#12 h Dec 25, 2020, 2:25 AM • 3 Y

Y by teomihai, eazy_math, guptaamitu1

Let $p\ge q\ge r\ge s$ . Observe that $(pq+rs)+(pr+qs)+(ps+qr)=\frac{(p+q+r+s)^2-\left(p^2+q^2+r^2+s^2\right)}2=30,$ but by Rearrangement, $pq+rs\ge pr+qs\ge ps+qr$ , so $pq+rs\ge10$ .

Note the following: $\begin{align*} (p+q)+(r+s)&=9\\ (p+q)^2+(r+s)^2&\ge41. \end{align*}$ Since $p+q\ge r+s$ , we must have $p+q\ge5$ .

Finally, $25\le(p+q)^2+(r-s)^2=21+2(pq-rs),$ so $pq-rs\ge2$ , as needed. Equality holds at $(p,q,r,s)=(3,2,2,2)$

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sqing

#13 h Mar 30, 2021, 1:03 PM

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Let $a, b, c,$ and $d$ be real numbers such that $a \geq b \geq c \geq d$ and $a+b+c+d = 13,a^2+b^2+c^2+d^2=43.$ Show that $ab-cd \geq 3 .$ Equality holds for $(4,3,3,3).$
2021 Philippine

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#14 h Apr 20, 2021, 7:36 PM • 1 Y

Y by teomihai

TheUltimate123 wrote:

Let $p\ge q\ge r\ge s$ . Observe that $(pq+rs)+(pr+qs)+(ps+qr)=\frac{(p+q+r+s)^2-\left(p^2+q^2+r^2+s^2\right)}2=30,$ but by Rearrangement, $pq+rs\ge pr+qs\ge ps+qr$ , so $pq+rs\ge10$ .

Note the following: $\begin{align*} (p+q)+(r+s)&=9\\ (p+q)^2+(r+s)^2&\ge41. \end{align*}$ Since $p+q\ge r+s$ , we must have $p+q\ge5$ .

Finally, $25\le(p+q)^2+(r-s)^2=21+2(pq-rs),$ so $pq-rs\ge2$ , as needed. Equality holds at $(p,q,r,s)=(3,2,2,2)$

" $(p+q)+(r+s)=9 (p+q)^2+(r+s)^2 \ge41.$ Since $(p+q\ge r+s )$ , we must have $(p+q\ge5).$ "

How do you get $p+q \geq$ ?

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#15 h May 1, 2021, 2:24 PM

Y by

Wizard_32 wrote:

who wrote:

Four real numbers $p$ , $q$ , $r$ , $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$ . Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2$ .

Assume that $\max\{pq,pr,ps,rq,rs,qs\}=pq,$ and assume on the contrary that $pq<rs+2.$ Since $pq+pr+ps+rq+rs+qs=60,$ hence $pq \geqslant 10.$ In particular, this means $2+rs>10 \implies rs>8>0.$ Then since $p^2,q^2,r^2,s^2$ are positive reals, hence by AM-GM
$\begin{align*} 8^2 < (rs)^2 \leqslant pqrs \leqslant \left( \frac{p^2+q^2+r^2+s^2}{4}\right)^2 \implies 8^2 \cdot 4^2 < 21^2 \end{align*}$ which is clearly false. $\square$

$pq+pr+ps+rq+rs+qs=60$ ; this is false, actually we have $(p+q+r+s)^2 = p^2 + q^2 + r^2 + s^2 + 2(pq + pr+ps+rq+rs+qs ) = 81 = 21 + 2(pq + pr+ps+rq+rs+qs )$ , and hence $pq+pr+ps+rq+rs+qs=30$ . Notice this implies $5 \leq pq$ , which means your solution is wrong.

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#16 h May 28, 2021, 1:16 PM

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dame dame

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#17 h Jul 1, 2021, 2:46 PM • 2 Y

Y by teomihai, centslordm

Here's a different solution with a weird substitution.
WLOG suppose that $p \geq q \geq r \geq s$ . I will show that $pq-rs \geq 2$ . Since $p+q+r+s=9$ , we can substitute
$\begin{align*} p&=\frac{9}{4}+k+x\\ q&=\frac{9}{4}+k-x\\ r&=\frac{9}{4}-k+y\\ s&=\frac{9}{4}-k-y, \end{align*}$ where $k,x,y\geq 0$ . Since $q \geq r$ , we require $2k \geq x+y$ . Using the substitution, the condition $p^2+q^2+r^2+s^2=21$ becomes
$\frac{81}{4}+\frac{9}{2}(k+x+k-x-k+y-k-y)+(k+x)^2+(k-x)^2+(k+y)^2+(k-y)^2=\frac{81}{4}+4k^2+2x^2+2y^2=21\implies 2k^2+x^2+y^2=\frac{3}{8}.$ Finally, the inequality $pq-rs\geq 2$ is equivalent to
$9k+y^2-x^2\geq 2$ Suppose now that we fix $k$ . Then it is clear that
$9k+y^2-x^2\geq 9k-4k^2,$ and since the inequality $9k-4k^2\geq 2$ holds for $\tfrac{1}{4} \leq k \leq 2$ , $pq-rs\geq 2$ holds in that range of $k$ as well. If $k>2$ , then
$2k^2+x^2+y^2\geq 2k^2>8>\frac{3}{8},$ so we cannot have $p^2+q^2+r^2+s^2=21$ in that case. Thus we can discard the case of $k>2$ . If $k<\tfrac{1}{4}$ , then we have
$\frac{3}{8}=2k^2+x^2+y^2<\frac{1}{8}+x^2+y^2 \implies x^2+y^2>\frac{1}{4} \implies (x+y)^2>\frac{1}{4} \implies x+y>\frac{1}{2}>2k,$ which contradicts $2k\geq x+y$ . Thus we can also discard $k<\tfrac{1}{4}$ , leaving only $k \in [\tfrac{1}{4},2]$ which we already proved $pq-rs\geq 2$ for. Thus we're done. $\blacksquare$

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#20 h Aug 28, 2021, 7:15 PM

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WLOG $p\ge q\ge r\ge s$ . We claim that $pq-rs \ge 2$ .
If $s\le 0$ , then, by Cauchy-Schwarz, $21 = p^2+q^2+r^2+s^2 \ge \frac{1}{3}(p+q+r)^2+s^2 \ge \dfrac{1}{3}(9)^2+s^2 \Rightarrow s^2\le -6$ which is clearly absurd. Hence, $p,q,r,s \in \mathbb{R}^+$ .
We have
$\begin{align*} &(9-q-r-s)^2+q^2+r^2+s^2=21 \\ \Leftrightarrow \quad &2q^2-2(9-r-s)q+(9-r-s)^2+r^2+s^2-21=0 \\ \Leftrightarrow \quad &q=\dfrac{2(9-r-s)\pm \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2} \end{align*}$ If $q=\dfrac{(9-r-s)+ \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2}$ , then
$q\le p = 9-q-r-s = q-\sqrt{42-(2r^2+2s^2+(9-r-s)^2)} \le q$ so $42-(2r^2+2s^2+(9-r-s)^2) = 0$ which will be included in the other case.
Therefore, we can assume that $q=\dfrac{(9-r-s)- \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2}$ . We will show that $r+s\le 4$ . Suppose, ftsoc, that $r+s > 4$ . We have
$\begin{align*} &\dfrac{(9-r-s)- \sqrt{42-(2r^2+2s^2+(9-r-s)^2)}}{2} \ge r \\ \Leftrightarrow \quad &3r^2+(2s-18)r+(s^2-9s+30)\ge 0 \\ \Leftrightarrow \quad &3\left(r+\dfrac{s-9}{3}\right)^2+\dfrac{2s^2-9s+9}{3}\ge 0 \end{align*}$
Claim: $\dfrac{3}{2}\le s \le \dfrac{9}{4}$

Proof: If $s< \dfrac{3}{2}$ , then $p+q+r+s \ge 3r+s > 3(4-s)+s > 12-2(\dfrac{3}{2}) = 9$ , a contradiction. So $s \ge \dfrac{3}{2}$ . Also, since $p+q+r+s=9$ , $s\le \dfrac{9}{4}.$ $\blacksquare$

By the claim, $2s^2-9s+9= (2s-3)(s-3)\le 0$ . We can now consider 2 possible cases.

Case 1: $r\ge\dfrac{9-s}{3}+\sqrt{\dfrac{-2s^2+9s-9}{9}}$
$9=p+q+r+s \ge 3r+s \ge (9-s)+3\sqrt{\dfrac{-2s^2+9s-9}{9}} +s =9+3\sqrt{\dfrac{-2s^2+9s-9}{9}}$ so $2s^2-9s+9=0 \Leftrightarrow s=3, \dfrac{3}{2}$ and $p=q=r$ . By claim, $s= \dfrac{3}{2}$ giving $p=q=r=\dfrac{5}{2}.$ Thus, $r+s = 4$ , a contradiction.

Case 2: $r\le\dfrac{9-s}{3}-\sqrt{\dfrac{-2s^2+9s-9}{9}}$
$s\le \dfrac{9-s}{3}-\sqrt{\dfrac{-2s^2+9s-9}{9}} \Leftrightarrow 2s^2-9s+10 \Leftrightarrow (2s-5)(s-2)\ge 0$ . By claim, $s\le 2$ .
However, $4-s < r \le\dfrac{9-s}{3}-\sqrt{\dfrac{-2s^2+9s-9}{9}} \Rightarrow 2s^2-7s+6>0 \Leftrightarrow (s-2)(2s-3)>0$ , which is a oontradiction since $\dfrac{3}{2}\le s\le2$ .

We therefore have that $r+s \le 4$ and
$\begin{align*} pq-rs &= (9-q-r-s)q-rs\\ &= \dfrac{81-((9-q-r-s)^2+q^2+r^2+s^2)}{2}+(r^2+s^2)-9(r+s) \\ &= 30 +(r^2+s^2)-9(r+s) \\ &= 22 + (r^2+4) +(s^2+4) -9(r+s) \\ &\ge 22 -5(r+s) \\ &\ge 2 \end{align*}$ as desired.

This post has been edited 1 time. Last edited by Afternonz, Aug 28, 2021, 7:21 PM

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#21 h Jan 22, 2022, 9:14 PM • 1 Y

Y by teomihai

Here's a different solution (which is a bit longer, though every step is easily motivated):

WLOG $p \le q \le r \le s$ . We will show $pq - rs \ge 2$ . Observe,
$\sum_{\text{sym}} pq = \frac{p^2 - 21}{2} = 30 \implies \sum_{\text{sym}} (p-q)^2 = 3 \cdot 21 - 2 \cdot 30 = 3 \qquad \qquad (1)$ Let $p-q = x, p - r = y, p -s = z$ . Observe $0 \le x \le y \le z$ . Then $p = \frac{p+x+y+z}{4}$ , thus
$\begin{align*} T &= pq - rs = p(p-x) - (p-y)(p-z) = p^2 - px - p^2 + p(y+z) - yz = p(y+z-x) - yz \\ &= \frac{(9 + x + y +z)(y+z-x)}{4} - yz = \frac{\left( y +z + \frac{9}{2} \right) - \left(x + \frac{9}{2} \right) }{4} - yz \\ &= \frac{ y^2 + z^2 + 9y + 9z + 2yz - x^2 - 9x }{4} - \frac{yz}{2} = \frac{9(y+z-x) + y^2 + z^2 - x^2}{4} - \frac{yz}{2} \\ &= \frac{9(y+z-x) + (z-y)^2 - x^2}{4} \ge \frac{9(y+z-x) - x^2}{4} \end{align*}$ So it suffices to show $S = 9(y+z-x) - x^2 \ge 8$ .

Claim: $x \le 1$ and $3z^2 + y - x \ge 3$ .
Proof: Using $(1)$ we have $3 \ge x^2 + y^2 + z^2 = 3x^2$ , giving $x \le 1$ . Also,
$\begin{align*} 3 - 3z^2 &= x^2 + y^2 + z^2 + (x-y)^2 + (y-z)^2 + (z-x)^2 - 3z^2 = 3(x^2 + y^2) - 2z(x+y) - 2xy \\ &= x(3x - 2z - 2y) + y(3y - 2z) \le x(3x - 4y) + y^2 = (y-x)(y-3x) \le y-x \end{align*}$ This proves our claim. $\square$

Now if $z \ge 1$ , then $S \ge 9z - x^2 \ge 9 - 1 = 8$ . Otherwise, if $z \le 1$ then
$\begin{align*} S \ge 9(3 + z - 3z^2) - 1 = 26 + 9z - 27z^2 = 8 + 9(2 + z - 3z^2) = 8 + 9(1-z)(2+z) \ge 8 \end{align*}$ This completes the proof. $\blacksquare$

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math90

#22 h May 10, 2022, 8:15 AM • 1 Y

Y by teomihai

Nothing new here.

WLOG assume that $p\ge q\ge r\ge s$ .

Note that
$pq+rs+pr+qs+ps+qr=\frac{(p+q+r+s)^2-(p^2+q^2+r^2+s^2)}{2}=30.$ Since $p\ge q\ge r\ge s$ we have $pq+rs=\max(pq+rs,pr+qs,ps+qr)$ . Hence $pq+rs\ge 10$ . This proves that
$(p+q)^2+(r+s)^2=(p^2+q^2+r^2+s^2)+2(pq+rs)\ge 21+2\cdot 10=41.$ Now let $t\doteqdot p+q$ . Then $t^2+(9-t)^2\ge 41$ , or equivalently $(t-4)(t-5)\ge 0$ . As $t\ge 4.5$ we obtain $t\ge 5$ . Therefore
$(p+q)^2\ge 25=p^2+q^2+r^2+s^2+4\ge p^2+q^2+2rs+4.$ This rearranges to $pq-rs\ge 2$ .

This post has been edited 1 time. Last edited by math90, May 10, 2022, 8:22 AM

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awesomeming327.

#23 h Jun 10, 2022, 11:19 PM • 2 Y

Y by teomihai, lian_the_noob12

Note that we have $pq+pr+ps+qr+qs+rs=30.$ Let $p\ge q\ge r\ge s$ and consider $pq-rs.$ Note that $pq+rs\ge pr+qs\ge ps+qr$ so $pq+rs\ge 10.$ Let $a=p+q$ and $b=r+s$ then $(p+q)^2\ge 4pq$ gives:
$a+b=9$ $a^2+b^2\ge 40$ Thus, $-(a-b)^2=(a^2+b^2+2ab)-2(a^2+b^2)\le -1$ so $a-b\ge 1.$ This implies $a\ge 5.$ Therefore, we have $(p+q)^2+(r-s)^2\ge 25.$ On the other hand $(p+q)^2+(r-s)^2=p^2+q^2+r^2+s^2+2(pq-rs)=21+2(pq-rs)$ so $pq-rs\ge 4$ as desired.

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DakuMangalSingh

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DakuMangalSingh

#24 h Aug 7, 2022, 12:13 PM • 1 Y

Y by teomihai

ISL Marabot Solve

WLOG, $p\geq q\geq r\geq s$ . We will prove, $pq-rs\geq 2$ .

FTSOC, let $pq-rs < 2$ .
Now $\sum_{sym} pq = \frac{(p+q+r+s)^2-(p^2+q^2+r^2+s^2)}{2}=30$ , So, by rearrangement inequality, $pq+rs\geq 10$ . If $rs < 4$ then $pq-rs= pq+rs - 2rs > 10-8=2$ , But as we assumed, this is not true. So, $rs\geq 4 \implies \frac{r^2+s^2}{2}\geq rs \geq 4\implies r^2+s^2\geq 8$ (By, AM-GM). Now, $p^2+q^2-r^2-s^2=\sum p^2 - 2(r^2+s^2)\leq 21-2\times 16=5$ .

Now let $p+q=x, r+s=9-x$ So, $\sum p^2 + 2(pq+rs)\geq 21+2\times 10=41 \implies (p+q)^2+(r+s)^2=(x^2)+(9-x)^2\geq 41$ $\implies (x-5)(x-4)\geq 0 \implies x\geq 5$ or $x\leq 4$ , Since $p\geq q\geq r\geq s$ and $\sum p = 9$ , so, $x=p+q\geq 5$ and $9-x=r+s\leq 4$

So, $(p+q)^2-(r+s)^2= p^2+q^2-r^2-s^2+2(pq-rs)\geq 5^2-4^2=9 \implies 2(pq-rs)\geq 9-(p^2+q^2-r^2-s^2)\geq 9-5=4$ $\implies pq-rs\geq2$ . But we guessed, $pq-rs<2$ . So, by contradiction, $pq-rs\geq 2$

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#25 h May 25, 2023, 12:00 PM • 1 Y

Y by teomihai

Solution

This post has been edited 3 times. Last edited by VicKmath7, May 25, 2023, 12:38 PM

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lian_the_noob12

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lian_the_noob12

#26 h Jun 9, 2023, 5:14 PM

Y by

$\color{green} \boxed{\textbf{SOLUTION}}$

We have,
$(p+q+r+s)^2=p^2+q^2+r^2+s^2+2(pq+pr+ps+qr+qs+rs)=81 \implies pq+pr+ps+qr+qs+rs=30$
Assume, $p \ge q \ge r \ge s$

We need to show, $pq-rs \ge 2$

By Rearrangement Inequality,
$pq+rs \ge pr+qs \ge ps+qr$
So, $pq+rs \ge 10$

Now, $(p+q)^2 + (r+s)^2=p^2+q^2+r^2+s^2+2(pq+rs) \ge 41$
$2[(p+q)^2+(r+s)^2]=[(p+q)+(r+s)]^2 + [(p+q)-(r+s)]^2 \implies -[(p+q)-(r+s)]^2=[(p+q)+(r+s)]^2-2[(p+q)^2+(r+s)^2] \le 81-82= -1$ So, $(p+q)-(r+s) \ge 1$
And, $p+q \ge 5$

$2(pq-rs)=p^2+q^2+r^2+s^2+2pq-2rs-21=(p+q)^2 + (r-s)^2 - 21 \ge 25+(r-s)^2-21 \ge 4 \implies pq-rs \ge 2 \blacksquare$

This post has been edited 5 times. Last edited by lian_the_noob12, Jun 9, 2023, 9:26 PM

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#27 h Jul 31, 2023, 4:42 AM • 1 Y

Y by teomihai

who wrote:

Four real numbers $p$ , $q$ , $r$ , $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$ . Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2$ .

This problem reminds me of several similar ones relating $a_1+...+a_n=k$ with $a_1^2+...+a_n^2=l^2$ and trying to maximize an element lol (for example in particular that AMSP test 3 Level 2 question or smth or that USAMO 1978 or smth like that
WLOG $p\ge q\ge r\ge s.$ We see that $pq+rs+pr+ps+qr+qs=30\implies pq+rs\ge 10$ by rearrangement ineq. Then note that we want to find maximum of rs so that we can subtract it. To do this, we want to maximize the sum r+s, so we are motivated to set t=r+s, otherwise it would be hard to show a maximum sum that also satisfies the second equality. Now, $t^2+(9-t)^2=p^2+q^2+2pq+r^2+s^2+2rs=41\implies (t-4)(t-5)\ge 0\iff 4\ge t=r+s,$ where the last step is because $r+s\le 9/2<5$ so it couldn't be bounded below by 5. Indeed, we see that maximum of rs=4 when r=s=2, and the inequality still holds since $pq+4\ge 10\iff pq\ge 6\implies pq-rs\ge 2. \blacksquare$

This post has been edited 1 time. Last edited by huashiliao2020, Jul 31, 2023, 4:42 AM

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#28 h Nov 29, 2023, 5:07 PM

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WLOG $p >= q >= r >= s$ . Then we aim to prove that $pq - rs >= 2$ .
We know that $r + s = 9 - p - q$ , and that $r^2 + s^2 = 21 - p^2 - q^2$
So, $r^2 + s^2 + 2rs = 81 + p^2 + q^2 - 18p - 18q - 2pq$ , or that
$2rs = 60 + 2p^2 + 2q^2 - 18p - 18q - 2pq$
$rs - pq = 30 + p^2 + q^2 - 9p - 9q - 2pq$
$pq - rs = 2pq - p^2 - q^2 + 9p + 9q - 30$
So, we have that $pq - rs >= 2$ is the same as $2pq - p^2 - q^2 + 9p + 9q - 30 >= 2$
Or, $9(p+q) >= 32 + (p-q)^2$
Clearly, $(p+q) >= \frac{9}{2}$ , so $9(p+q) >= \frac{81}{2}$
So, we have $\frac{17}{2} >= (p-q)^2$ .
The maximum value of ${p,q,r,s}$ is $3$ , and the minimum value is $\frac{3}{2}$ , so $p-q$ cannot exceed $3/2$ , or $(p-q)^2$ cannot exceed $9/4 < 17/2$ .

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#29 h Apr 29, 2024, 4:29 PM • 2 Y

Y by teomihai, Safal

Assume WLOG $p \ge q \ge r \ge s$ . We aim to show $pq-rs \ge 2$ . Notice
$\sum_{\text{cyc}} ab = \tfrac 12(9^2-21) = 30 \implies pq+rs \ge 10$
by Rearrangement inequality. We also have
$(p+q)^2+(r+s)^2 = 21+2(pq+rs) \ge 41 \implies p+q \ge 5$
from the first condition and $p+q \ge r+s$ . Finally,
$25 \leq (p+q)^2+(r-s)^2 = 21+2(pq-rs) \implies pq-rs \ge 2. \quad \blacksquare$

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OronSH

#30 h Aug 27, 2024, 5:04 AM

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Make the substitution $(p,q,r,s)=(a+2,b+2,c+2,d+2)$ such that $a+b+c+d=1,a^2+b^2+c^2+d^2=1.$ WLOG $a\ge b\ge c\ge d.$ If $c,d\le 0$ then $c+d\le 0.$ If $d>0$ then $1>a,b,c,d>0$ but $a>a^2$ in this range, impossible.

Now suppose $a\ge b\ge c\ge 0\ge d$ and $c+d>0.$ Notice we can get $ab+bc+cd+da+ac+bd=(a+b)(c+d)+ab+cd=0,$ and $a+b\ge c+d>0$ so $ab+cd<0.$ Now $ab$ is positive and $cd$ is negative so $|ab|<|cd|.$ Since $|b|>|c|$ we have $|a|<|d|$ so $a+d<0.$

Now $a+d<0\implies b+c>1\implies b^2+c^2>\tfrac12\implies a^2+d^2<\tfrac12\implies a^2<\tfrac14\implies a<\tfrac12\implies b,c<\tfrac12\implies b+c<1,$ contradiction. Thus $c+d\le 0.$

We have shown $c+d\le 0$ always, so $a+b\ge 1$ and $p+q\ge 5.$ Then $(p+q)^2+(r-s)^2=21+2(pq-rs)\ge 25,$ done.

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#31 h Sunday at 11:00 PM • 1 Y

Y by OronSH

had a bad bashy method that i couldn't finish so here goes the better solution with some of what i thought could be nice motivation.

also missed @above method which also fits into this idea of "doing some algebra-y tricks" i guess i should try putting that idea into my toolkit ig

Assume that $p\ge q\ge r\ge s$ . Evidently the signs of $(p,q,r,s)$ must be one of $(+,+,-,-)$ , $(+,+,+,-)$ , or $(+,+,+,+)$ , as $(-,-,-,-)$ obviously fails and having only $p$ positive would cause $p^2>21$ .

In the latter two cases, the optimal permutation that maximizes the value of $ab-cd$ is evidently the original order of $(p,q,r,s)$ , so that it remains to prove $pq-rs\ge 2$ . In the first case, this is also true; we will ultimately have to subtract two numbers of the same sign, so we should make the magnitude of one product as large as possible, which is achieved with $pq$ (as these two have a magnitude of sum which is larger than the magnitude of sum of $r$ and $s$ ).

Hence it remains to prove only that $pq-rs\ge 2$ , so that we have dropped the "exists a permutation" condition---effectively we just have a single concrete statement to prove. It's technically "stronger" in the sense that it may not be the best difference of two products that we could have picked, but it turns out it works.

Okay this is where I started to bash. But following the idea of just doing a little amount of algebra-y tricks, let's realize that $pq$ is closely related to $p^2+2pq+q^2$ . In light of this, let's consider
$(p+q)^2+(r-s)^2=p^2+q^2+r^2+s^2+2pq-2rs=21+2pq-2rs$ which is especially nice as it uses a condition given in the problem (sum of the squares) while also using $pq-rs$ , the expression of interest.

Here's the sort of sketchy part. The way I'm going to look at this is from the perspective of "brainstorming"---it's not the kind of thing that you know is going to work right off the bat, but it's an idea that you have to brainstorm and then as you brainstorm other ideas you realize that the idea I'm about to talk about proves useful.

We know that $\sum{pq}=30$ from squaring the first given condition and subtracting the second. Then, seeing that we are working with $pq$ minus $rs$ , it is quite reasonable and even important if we're keeping track of things that could be useful in the future to note that $pq+rs\ge 10$ by Rearrangement (which I need to look at along with other inequality theorems and inequalities in general).

At this point, the brainstorming idea of using $(p+q)^2$ is back. This time, we still want the final expression to have $p^2+q^2+r^2+s^2$ , but we also want $pq$ to be ADDED to $rs$ . When we realize that the resulting expression of
$(p+q)^2+(r+s)^2=21+2pq+2rs\ge 41$ also uses $r+s$ , it becomes even better: we can write $t=p+q$ to get
$t^2+(9-t)^2=2t^2-18t+81\ge 41\implies (t-4)(t-5)\ge 0$ and that means $t=p+q\ge 5$ , since we also have $p+q\ge r+s$ .

Now we're just done: we have brainstormed all of this nice information, and it comes together in the following:
$25\le (p+q)^2\le (p+q)^2+(r-s)^2=21+2pq-2rs\implies 2\le pq-rs.$ yahoo that was actually quite fun to write okay and we're done and i feel like i sort of learned something here about knowing how to brainstorm even and committing to thinking of just any ideas that might be useful even if i don't see a path to the solution or progress immediately which is nice hopefully that serves me well on usamo dunno tho locked in next question here goes ig guh sus yoink $\blacksquare$

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