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.
\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}
Let us try to minimize $$\varphi = a y + b x - ac - x z$$
We suppose that $a,b,c,x,y,z$ are given, satisfying the required conditions.
Without any computation one can decrease $\varphi$ by decreasing $b$ and $y$ by the same amount. Indeed, it doesn't change the condition $a+b+c = x+y+z$ and it decreases $\varphi$ because $a+x\ge 0$. Hence we can suppose that $\boxed{b+y=c+z}$.
With the same argument, one can decrease $\varphi$ by decreasing $a$ and $x$ by the same amount, because $b+y\ge c+z$. We can do this until $\boxed{a+x=b+y}$.
Let $2 m:=a+x = b+y=c+z$. One has $6m = a+b+c+ x+y+z=2 (a+b+c)$ hence $\boxed{a+b+c = 3m}$ and $x + y+ z = 3m$. One has $\boxed{x = 2m-a}$, $\boxed{y = 2m-b}$, $\boxed{z = 2m-c}$
Replacing these values into $\varphi$, leads to $$\boxed{\varphi = 2 (m-a)^2\ge 0}$$
Indeed, $$ay+bx=a(2m-b)+b(2m-a)=2m(a+b) - 2ab=2m(3m-c)-2ab$$ and $$-ac-xz = -ac-(2m-a)(2m-c) = -4m^2+2m(a+c)-2ac$$ hence $$\varphi=2m^2+2ma-2a(b+c) = 2m^2+2ma-2a(3m-a)=2(m-a)^2$$
