MHB Can you solve this week's POTW #474: Four real numbers with unique permutations?

[anemone]

Here is this week's POTW:

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

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[anemone]

No one answer POTW#474.

Spoiler

However, you can refer to the suggested solution as follows:

WLOG, let $a\ge b \ge c \ge d$. Assume that max ${ab,\,ac,\,ad.\,cb,\,cd,\,bd}=ab$ and assume on the contrary that $ab<cd+2$.

Since $ab+ac+ad+cb+cd+bd=\dfrac{(a+b+c+d)^2-(a^2+b^2+c^2+d^2)}{2}=30$, $ab+ac+ad+cb+cd+bd=60$. Hence $ab\ge 10$. In particular, this means $2+cd>10$, which gives $cd>8>0$. By the AM-GM inequality,

$8^2<(cd)^2\le abcd \le \left(\dfrac{a^2+b^2+c^2+d^2}{4}\right)^2$

This implies $32^2<21^2$, which is clearly false.

That suggest $ab-cd>2$ is true.