Query about an algebraic inequality involving $(a-b)^p$ I want to know if there exists any inequality of the type $(a-b)^p \geq C(a^p -b^p)$ or $(a-b)^p \leq C(a^p -b^p),$ where $a>0,\, b>0,$ $C>0$ is a constant and $0<p<1.$ I am aware of the inequality of the type $(a+b)^p \geq C (a^p+b^p)$ where $a,b \geq 0$ $C>0$ is a constant and $0<p<1.$

It seems the following. Divide both sides by $a^p$ and substitute $x=\frac ba>0$. The inequality reduces to $(1-x)^p\,?\, C(1-x^p)$. Since the case $x>1$ may cause $(1-x)^p$ undefined, we assume that $x<1$ (the case $x=1$ yields the equality). Next, when $x$ tends to $1$ then Tailor series for the function $(1-y)^p$ yields $1-x^p=1-(1-(1-x))^p=p(1-x)+o((1-x)^2)=o((1-x)^p)$, so only the case when "$?$" is $\ge$ is possible. When $x$ tends to $0$ then both $(1-x)^p$ and $(1-x^p)$ tend to $1$, so $C\ge 1$. The case $C=1$ works, because $(1-x)^p+x^p\ge 1-x+x=1$. 
Answer: $(a-b)^p\ge a^p-b^p$.
