Let $0\leq a,b\leq 1$ and $1<p<\infty$, then prove the following inequality: $$ |a^p-b^p|\leq p(a^p+b^p)^\frac{p-1}{p}|a-b|. $$ Thanks in advance.
I tried to prove in the following way and hope this is correct: Kindly inform me.
Let $f(x)=x^p$, $x\in[0,1]$. Then by the MVT, we know for some $t\in(0,1)$ $$ f(b)-f(a)=(b-a)f'(a+tb)=(b-a)p(a+tb)^{p-1}\leq p|a-b|(a+tb)^{p-1}\leq p|a-b|(a+b)^{p-1}\leq p2^{p-1}|a-b|(a^p+b^p)^\frac{p-1}{p}, $$ where the last inequality above holds due to the fact that $$ (a+b)^{p}\leq 2^{p}(a^p+b^p). $$
But I want to constant to be $p$ in place of $p2^{p-1}$ on the right-hand side. Can somebody please help me with the same? Thanks.
I do not know how to finish with your approach but I give you my proof just using some simple calculations.
Let $x=\frac{a}{b}$ and assume that $x\geq 1$. The inequality is equivalent to
$$x^p-1 \leq p(x^p+1)^{\frac{p-1}{p}}(x-1)$$
Define $f(x)=p(x^p+1)^{\frac{p-1}{p}}(x-1)-x^p+1$ and note that $f(1)=0$.
If my computation is right, then we have
\begin{align} f'(x)&=p(x^p+1)^{\frac{p-1}{p}}+p(p-1)(x-1)(x^p+1)^{-\frac{1}{p}}-px^{p-1}\\ &\geq p(x^p+1)^{\frac{p-1}{p}}-px^{p-1}\\ &>0. \end{align} So $f(x)$ is increasing, then $f(x)\geq f(1)=0$.