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For $a,b \in\mathbb{R}$, some $C_\alpha>0$, \begin{align*} &\big ||a+b|^\alpha - |a|^\alpha \big |\leq|b|^\alpha,&\quad \text{for }\alpha\leq 1,\\\\ &\big ||a+b|^\alpha - |a|^\alpha \big |\leq C_\alpha\big(|a|^{\alpha-1}+|b|^{\alpha-1} \big)|b|, &\quad \text{for }\alpha>1 \end{align*} I kindly ask for hints how to prove these inequalities hold true. I could just notice that when $|a+b|^\alpha\geq|a|^\alpha$, then $|a+b|^\alpha\leq|a|^\alpha+|b|^\alpha$, but unfortunately I have no idea of how to e.g. apply the knowledge about the polynomial/exponential function $c^\alpha=e^{\alpha\log c}$ or something else.

Thank you.

Shaun
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Alexey
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  • In general $|a+b|^p\leq(|a|+|b|)^p\leq 2^p \max(|a|^p,|b|^p)\leq2^p(|a|^p+|b|^p)$. This also has been addressed before in MSE. – Mittens Jul 24 '21 at 22:53
  • I proved the first one but the second one is still puzzling me. Can we prove it manipulating only with the left side not taking the right one into account? – Alexey Jul 25 '21 at 20:41

1 Answers1

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This addresses the second inequality stated by the OP. The following general inequality will be useful: $$(1+t)^\alpha\leq 2^\alpha(1+t^\alpha)$$ for all $t\geq0$ and $\alpha\geq0$.

For $p\geq1$, simple integration yields $$(1+x)^p-1=\int^x_0p(1+t)^{p-1}\,dt\leq p\int^x_02^{p-1}(1+t^{p-1})\,dt\leq p2^{p-1}(x+x^p)$$ Hence $$(1+x)^p\leq 1+p2^{p-1}(x+x^p)$$ Consequently, for $a>0,b\geq0$ $$\begin{align} (a+b)^p&=a^p\Big(1+\frac{b}{a}\Big)^p\leq a^p\Big(1+p2^{p-1}\big(\frac{b}{a}+\frac{b^p}{a^p}\big)\Big)\\ &=a^p+ p2^{p-1}\big(a^{p-1}b+b^p\big) \end{align} $$

Mittens
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