<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] Given that $x+y+z=0$, we have two of them to be of different sign. Without loss of generality, let us assume $x$ and $y$ have different sign, i.e., $xy\leq0$.
We have $z=-(x+y)$ and hence
\begin{align}
a^2yz + b^2zx + c^2xy & = c^2xy - (x+y)(a^2y+b^2x)\\
& = xy(c^2-a^2 - b^2) - a^2x^2 - b^2y^2\\
& = xy(c^2-a^2-b^2+2ab) - (ax+by)^2\\
& = xy (c^2-(a-b)^2) - (ax+by)^2\\
& = xy(c+a-b)(c-a+b) - (ax+by)^2 \leq 0
\end{align}
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