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We know the following inequality, \begin{equation*} (a+b)^{\alpha}\leq a^{\alpha} + b^{\alpha},\quad a,b \geq 0, \quad \alpha \in [0,1]. \end{equation*} Is it possible to write the following elementary inequality. If so, how we know.

\begin{equation*} (a+b)^{-\alpha}\leq a^{-\alpha} + b^{-\alpha},\quad \text{if}\quad a,b > 0,\quad \text{and} \quad \alpha >1, \end{equation*}

Thanks in advance.

M.cadirci
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1 Answers1

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$\dfrac{1}{(a+b)^{\alpha}}\leq\dfrac{1}{a^{\alpha}}+\dfrac{1}{b^{\alpha}}$ if and only if $1\leq\dfrac{(a+b)^{\alpha}}{a^{\alpha}}+\dfrac{(a+b)^{\alpha}}{b^{\alpha}}$ if and only if $1\leq\left(\dfrac{a+b}{a}\right)^{\alpha}+\left(\dfrac{a+b}{b}\right)^{\alpha}$ if and only if $1\leq\left(1+\dfrac{b}{a}\right)^{\alpha}+\left(1+\dfrac{a}{b}\right)^{\alpha}$.

But we see that $1+b/a>1$ and $1+a/b>1$ so $\left(1+\dfrac{b}{a}\right)^{\alpha}+\left(1+\dfrac{a}{b}\right)^{\alpha}>2$ actually.

user284331
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  • Thanks for your answer, but I am a little confused. As a result, can we write this, $$(a+b)^{-\alpha}\leq a^{-\alpha} + b^{-\alpha}$$, – M.cadirci Oct 27 '19 at 16:40
  • Yes, not only that, we can even write $2/(a+b)^{\alpha}<a^{-\alpha}+b^{-\alpha}$. – user284331 Oct 27 '19 at 16:48