How prove this inequality $\sum\limits_{cyc}\frac{x^a\ln{x}}{(x^a+y+z)^2}\ge 0$

I have a proof that applies for $a\geq -1$.

We start by making the change of variables $X = \exp(x)$ etc. The condition $xyz\geq 1$ becomes $X+Y+Z\geq 0$ and the ineqality we want to prove reads

$\sum_{\rm cyc}\frac{X\exp(aX)}{(\exp(aX) + \exp(Y) + \exp(Z))^2} \geq 0$

We combined the three terms to form a single fraction. The denominator is a product of squares and always positive. The nominator will consist of the three terms $Xf_1 + Yf_2 + Zf_3$

where

$f_1 = \exp(aX)(\exp(X)+\exp(aY)+\exp(Z))^2(\exp(X)+\exp(Y)+\exp(aZ))^2$

and $f_2,f_3$ are given by the same expression with a cyclic permutation of the variables.

To proceed we want to expand this product into single exponential terms and then use the approximation $\exp(x) > 1 + x$ on each of the terms. To avoid doing this manually we can use a trick: the approximation $\exp(x) > 1 + x$ implies $f_1 > T(f_1)$ where $T(f_1)$ is the first order Taylor series of $f_1$ around $(X,Y,Z)=(0,0,0)$

$T(f_1) = f_1(0,0,0) + \left.\frac{df_1}{dX}\right|_{(0,0,0)}X + \left.\frac{df_1}{dY}\right|_{(0,0,0)}Y + \left.\frac{df_1}{dZ}\right|_{(0,0,0)}Z$

which reads

$T(f_1) = 27(3 + (3a+4)X + 2(a+1)(Y+Z))$

and similar for $f_2,f_3$. Summing up for the three terms we find that the numerator satisfies

$Xf_1 + Yf_2 + Zf_3 \geq 27(2\alpha^2(a+1) + 3\alpha + (a+2)(X^2+Y^2+Z^2)) \geq 0$

at least for $a\geq -1/2$ since $\alpha \equiv X+Y+Z \geq 0$.

Equality is found only for $X=Y=Z=0$ or in terms of the original variables: $x=y=z=1$.