<s>[INST] prove that $a^2yz+b^2zx+c^2xy\leq 0$ for all real $x,y,x$ such that $x+y+z=0$ prove that $a^2yz+b^2zx+c^2xy\leq 0$ for all real $x,y,x$ such that $x+y+z=0$
  Given that $a,b,c$ are positive real numbers such that the sum of any two is greater than third.
 We have $x+y+z=0\implies x+y=-z\implies xz+yz=-z^2$
What will be the next?
 [/INST] We need to prove that $$-a^2y(x+y)-b^2x(x+y)+c^2xy\leq0$$ or
$$b^2x^2+(a^2+b^2-c^2)xy+a^2y^2\geq0$$
and since $b>0$, it's enough to prove that
$$(a^2+b^2-c^2)^2-4a^2b^2\leq0$$ or
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)\geq0$$
and we are done!
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