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

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 $pq-rs\geq 2$.

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 $(a,b,c,d) = (3,2,2,2)$)

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.

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.

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.

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 $(p,q,r,s) = (2,2,2,3)$.

nice solutions!

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.

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. $\square$

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 \((p,q,r,s)=(3,2,2,2)\)

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

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 $ ?

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.

dame dame

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. $\blacksquare$

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.

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. $\blacksquare$

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 $pq-rs\ge 2$.

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.

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, $pq-rs\geq 2$

$\color{green} \boxed{\textbf{SOLUTION}}$ 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, $p+q \ge 5$ $$2(pq-rs)=p^2+q^2+r^2+s^2+2pq-2rs-21=(p+q)^2 + (r-s)^2 - 21 \ge 25+(r-s)^2-21 \ge 4 \implies pq-rs \ge 2 \blacksquare$$

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 $$pq+4\ge 10\iff pq\ge 6\implies pq-rs\ge 2. \blacksquare$$

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 $9/4 < 17/2$.

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, \[25 \leq (p+q)^2+(r-s)^2 = 21+2(pq-rs) \implies pq-rs \ge 2. \quad \blacksquare\]

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.