Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. Prove that $abcd > a + b + c + d + 8$.

Question

Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. How to prove that $abcd > a + b + c + d + 8$? I've tried using AM-GM inequality but with no outcome. The problem was taken from 57-th Belarusian Mathematical Olympiad available here

score 3 · Accepted Answer · answered Mar 16 '17 at 18:48

It suffices to check the case $a, b, c, d>0$. Indeed, $abcd>0$, and so the problem becomes strictly stronger upon the transformation $(a, b, c, d)\to (|a|, |b|, |c|, |d|)$ if any of the variables are negative.

Now let $\frac{a+b+c+d}{4}=t$. Then $t^4\ge abcd>a^2+b^2+c^2+d^2\ge 4t^2$ due to AM-GM and Cauchy Schwarz, so $t> 2$ since $t>0$. Then we have: $$abcd>a^2+b^2+c^2+d^2\ge 4t^2=2t^2+2t^2>4t+8$$

Hence $abcd>4t+8$, which is exactly what we need to show.

Real numbers $a$, $b$, $c$ and $d$ satisfy the inequality $abcd > a^2 + b^2 + c^2 + d^2$. Prove that $abcd > a + b + c + d + 8$.

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