<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] There is also the following way.
Since for $\prod\limits_{cyc}(a+b-c)\leq0$ the inequality is true,
it's enough to prove our inequality for $\prod\limits_{cyc}(a+b-c)>0.$
Now, if $a+b-c<0$ and $a+c-b<0$ so $a+b-c+a+c-b<0,$
which is a contradiction.
Thus, it's enough to assume that $z=a+b-c>0$, $y=a+c-b>0$ and $x=b+c-a>0$
and we need to prove that
$$\left(\left(\frac{y+z}{2}\right)^2+\left(\frac{x+z}{2}\right)^2\right)^2\geq(x+y+z)xyz$$ or
$$(x^2+y^2+2z^2+2yz+2xz)^2\geq16(x+y+z)xyz,$$ which is true by AM-GM twice:
$$(x^2+y^2+2z^2+2yz+2xz)^2\geq(2xy+2z^2+2yz+2xz)^2=$$
$$=4(xy+z(x+y+z))^2\geq4\left(2\sqrt{xyz(x+y+z)}\right)^2=16(x+y+z)xyz.$$
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