Let $a>b>0$ such that $a+b=2$ and $\alpha\geq 1$ then we have : $$a^{ab}-b^{ab}\leq \alpha(a^{\frac{ab}{\alpha}}-b^{\frac{ab}{\alpha}})$$
My try :
The problem is equivalent to : $$\frac{a^{ab}-b^{ab}}{ab\ln(a)-ab\ln(b)}\leq \frac{(a^{\frac{ab}{\alpha}}-b^{\frac{ab}{\alpha}})}{\frac{ab}{\alpha}\ln(a)-\frac{ab}{\alpha}\ln(b)}$$
Hence it have the form (with $f(x)=e^x$):
$$\frac{f(ab\ln(a))-f(ab\ln(b))}{ab\ln(a)-ab\ln(b)}\leq \frac{f(\frac{ab}{\alpha}\ln(a))-f(\frac{ab}{\alpha}\ln(b))}{\frac{ab}{\alpha}\ln(a)-\frac{ab}{\alpha}\ln(b)}$$
But after this I don't know what to do...
Thanks in advance.
Since $a>b>0$ and $a+b=2$, we have $a > 1$ and $b<1$, so $c:=a^{ab}>1>b^{ab}=:d$.
Then,
$$\frac{\mathrm{d}}{\mathrm{d}\alpha} (\alpha(c^{1/\alpha}-d^{1/\alpha}) = c^{1/\alpha}-d^{1/\alpha} + \frac{1}{\alpha}(c^{1/\alpha}\ln c-d^{1/\alpha}\ln d) $$
and $c^{1/\alpha}\ln c \geq 0$, $-d^{1/\alpha}\ln d \geq 0$.