It's a follow up of Prove that $f(x,y)+f(y,z)\ge f(x,z)$ where $f(x,y)=\sqrt{x\ln x+y\ln y-(x+y)\ln(\frac{x+y}2)}$ :
I don't give the details but I show using second derivative :
if we have for $a,b,c,x>0$ $$F(x)=g(x,abx/c)-g(x,xa)+g(abx/c,xa)$$
Where :
$$g^2(x,y)=x\ln x+y\ln y-(x+y)\ln\frac{x+y}{2}$$
Then we have $x>0$ :
$$F''(x)\leq 0$$
So by Jensen's inequality we have $x,y>0$:
$$2F\left(\frac{x+y}{2}\right)\geq F(x)+F(y)\tag{I}$$
Now I ask :
Can we show $(I)$ using integral ? (As possible answer we can use Hermite-Hadamard inequality for convex/concave function ).
Have you an alternative proof for $(I)$ as I already answer simply the first question ?
Some details :
Let $0<x<1,a>1$ then the function :
$$f\left(x\right)=\sqrt{x\ln\left(x\right)-\left(x+1\right)\ln\frac{x+1}{2}}+\sqrt{a\ln\left(a\right)-\left(a+1\right)\ln\frac{a+1}{2}}-\sqrt{x\ln\left(x\right)+a\ln\left(a\right)-\left(x+a\right)\ln\frac{x+a}{2}}$$
is convex so we have :
$$f(x)\geq \left(x-1\right)\cdot\lim_{x\to 1^{-}}f'\left(1\right)>0$$
If $1<x<a$ then $f(x)$ is concave so we have :
$$f(x)>\left(x-1\right)\cdot\lim_{x\to a^{-}}\frac{\left(f\left(1\right)-f\left(x\right)\right)}{1-x}>0$$
If $x>a>1$ then we have : $$f'(x)<0$$ and $$\lim_{x \to \infty }f(x)>0$$
