<s>[INST] Elementary Inequality from a local olympiad test The problem I'm stuck with is the following :

Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$a+x \ge b+y \ge c+z \ge 0 $ and $ a+ b+c = x+y+z$
Prove that $ay+bx \ge ac+xz$

As easy as it sounds, I didn't achieve any significant progress after several tries.
Any advice, hint would be very appreciated. Thanks.
 [/INST] \begin{align}
(ay+bx) - (ac+xz) &= a(\color{red}{y-c}) + x(b-z) \\
&= a(\color{red}{a+b - x -z}) + x(b-z)\\
&= a(a-x) + (a+x)(b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a^2-x^2) + (a+x)(b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(a-x+2b-2z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(\color{red}{a+b-x-z}+b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(\color{red}{y-c} + b-z)\\
&= \tfrac12(a-x)^2 + \tfrac12(a+x)(b+y-c-z)\\
&\geqslant 0
\end{align}
</s>