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How to prove the following inequality?

$$|x^{\frac{1}{\alpha}}\log(x+1)-y^{\frac{1}{\alpha}}\log(y+1)|\geq |x-y|^{\frac{1}{\alpha}}\log(|x-y|^{\frac{1}{\alpha}}+1)$$ for non-negative $x$ and $y$ and $\alpha\in [0,1].$

Any help would be appreciated.

Thank you.

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Not exactly, I made a visual diagram to show it as following, in which the black region is the values of the right hand while the blue region is the values of the left hand. It is apparently that your inequality can not hold. Maybe you should modify the desired inequality as the follows: $$\left|x^{\frac{1}{\alpha}}\log(x+1)-y^{\frac{1}{\alpha}}\log(y+1)\right|\geq |x-y|^{\frac{1}{\alpha}}\log(|x-y|+1).$$enter image description here

And if we modify the inequality as above, then things become much simpler. First set $$f(x)=x^{\frac{1}{\alpha}}\log(x+1),$$ then we only have to verify the following $$f(|x-y|)\leq |f(x)-f(y)|.$$ Note that it is clearly that $f$ is a strictly convex function, that's the finnal key point which verifies the inequality. While there exist some ambiguities, let me give a detail statement. Assume $x>y$, we can set $x-y=t>0$. If $t<y$,then by the convexity of $f$, we get $$\frac{f(x)-f(y)}{x-y}>\frac{f(y)-f(t)}{y-t}>\frac{f(t)-f(0)}{t},$$ which implies $f(x)-f(y)>f(t)=f(x-y)$. And if $y<t<x$, then also by the convexity we have $$\frac{f(x)-f(y)}{x-y}>\frac{f(t)-f(y)}{t-y}>\frac{f(t)-f(0)}{t},$$ we also get $f(x)-f(y)>f(t)=f(x-y)$.

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  • $\begingroup$ I'm sorry, I don't get how to use convexity to prove $f(x-y)\leq f(x)-f(y)$ if, for example, we assume $x\geq y.$ $\endgroup$ Sep 23 '19 at 6:08
  • $\begingroup$ I have made a completion as above. $\endgroup$
    – mathon
    Sep 23 '19 at 7:39

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