I'm interested by the following problem :
Let $x,y,z>0$ such that $x\geq y \geq z$ and $x-y=y-z=0.5(x-z)$ then we have : $$f(x,y,z)=\Big(x^xy^yz^z\Big)^{\frac{1}{x+y+z}}+(xyz)^{\frac{1}{3}}\leq x+z$$
Since it's homogeneous we have : $$f(\alpha x , \alpha y , \alpha z)=\alpha f(x,y,z)$$ We put $\alpha=\frac{1}{y}$ and $u=\frac{x}{y}$ and $v=\frac{z}{y}$ we get : $$f(x,y,z)=yf(u,1,v)\leq y(u+v)$$ and the condition is equivalent to $u+v=2$ so we have the following equivalent for $1\leq v \leq 2$ : $$\Big(v^v(2-v)^{(2-v)}\Big)^{\frac{1}{3}}+(v(2-v))^{\frac{1}{3}}\leq 2$$
Wich is true graphically otherwise I try power series but it's too boring and derivatives also .
We can generalize this easily but it's an other subject .
I think futhermore that there is consequences in information theory .
My question is :
How to prove it elegantly ?
Edit : the link with Information theory :
The inequality is equivalent to :
$$\Big(x^xy^yz^z\Big)^{\frac{1}{x+y+z}}+y+(xyz)^{\frac{1}{3}}\leq x+y+z$$
Or if we divide by $x+y+z$ :
$$\Big((\frac{x}{x+y+z})^{\frac{x}{x+y+z}}(\frac{y}{x+y+z})^{\frac{y}{x+y+z}}(\frac{z}{x+y+z})^{\frac{z}{x+y+z}}\Big)+(\frac{y}{x+y+z})+(\frac{x}{x+y+z}\frac{y}{x+y+z}\frac{z}{x+y+z})^{\frac{1}{3}}\leq 1$$
So if we put $a=\frac{x}{x+y+z}$ and $b=\frac{y}{x+y+z}$ and $c=\frac{z}{x+y+z}$ we get :
$$a^ab^bc^c+b+(abc)^{\frac{1}{3}}\leq 1$$ With $a+b+c=1$ and $2b=a+c$
Or : $$a^ab^bc^c\leq 1-b-(abc)^{\frac{1}{3}}$$ If we take the log on both side we have :
$$a\log(a)+b\log(b)+c\log(c)\leq \log(1-b-(abc)^{\frac{1}{3}})$$
Wich is in link with Shannon's entropy .
Many thanks