<s>[INST] Olympiad inequality proof issue Prove that $(a^2+b^2)^2\geq(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \forall \ a,b,c\in\mathbb{R^+} $.
I, forgetting to consider whether $a_1$ and $a_2$ are strictly non-negative (don't think they are), found a proof (almost) using the AM-GM inequality with $a_{1}=(a+b+c)(a+b-c)=(a+b)^2-c^2$ and $a_2=(b+c-a)(c+a-b)=c^2-(a-b)^2$, leading to (after manipulation):
$4a^2b^2 \geq (c^2-(a-b)^2)((a+b)^2-c^2)=(a+b+c)(a+b-c)(b+c-a)(c+a-b)$
$4a^2b^2\leq(a^2+b^2)^2$, clearly, so initial inequality proven.
However, I just realised AM-GM only holds for non-negative reals, and my $a_1$ and $a_2$ are both non-negative only if $b-a\leq c\leq b+a$.
Is there another way to prove this using AM-GM, or an extension of the same idea that covers the cases where $c\gt b+a \gt b-a$ or $b+a \gt b-a \gt c$? Also, why does the proof so cleanly yield the result?
 [/INST] So your proof works fine when $a+b\geq c, b+c\geq a, c+a\geq b$. Note that, at most one term among $a+b-c,b+c-a,c+a-b$ can be negative otherwise, one among $a,b,c$ will become negative. Thus, when exactly one term among $a+b-c,b+c-a,c+a-b$ is negative, RHS of inequality becomes negative. However, LHS is always non-negative and so, the inequality follows.
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