Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.3 National olympiad Prove it

88. (Provided by Pham kim kung) Let $a, b, c, d \in \overline{\mathbf{R}^{-}}$ , and $a+b+c+d=4$ , then
$16+2 a b c d \geqslant 3(a b+a c+a d+b c+b d+c d)$

Equality holds if and only if $a=b=c=d=1$ , or one of $a, b, c, d$ is zero and the other three are equal to $\frac{4}{3}$ .

Solution

88. Proof: Let
$\begin{array}{c} s_{1}=a+b+c+d \\ s_{2}=a b+a c+a d+b c+b d+c d \\ s_{3}=b c d+a c d+a b d+a b c \\ s_{4}=a b c d \end{array}$

Then the original inequality to prove is
$s_{1}^{4}+32 s_{4} \geqslant 3 s_{1}^{2} s_{2}$

Inequality (1) can be derived from the following:
$\begin{aligned} s_{1}^{4}-3 s_{1}^{2} s_{2}+32 s_{4}= & \left(s_{1}^{4}-4 s_{1}^{2} s_{2}+9 s_{1} s_{3}-16 s_{4}\right)+ \\ & \left(s_{1}^{2} s_{2}-4 s_{2}^{2}+3 s_{1} s_{3}\right)+4\left(s_{2}^{2}-3 s_{1} s_{3}+12 s_{4}\right) \geqslant \\ & 0 \end{aligned}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.