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I've been so far trying to show:

$(\frac{a^2}{a^2 + b^2})^{p/2} + (\frac{b^2}{a^2 + b^2})^{p/2} \le 1.$

Also, it holds true that $\frac{a^2}{a^2 + b^2} \le 1$ and $\frac{b^2}{a^2+b^2}\le 1.$

I'm struggling with how to tie all of this together to get the desired result. Thanks for your help.

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3 Answers 3

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(Rather complete) hint: Using the binomial theorem ( https://en.wikipedia.org/wiki/Binomial_theorem ) on the right hand side, you find that every term of the right-hand side is positive. Leaving away certain terms yields smaller values. If you leave away the mixed terms, only $a^p + b^p$ will remain there!

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Just expand $$(a^p+b^p)^2=a^{2p}+b^{2p}+2a^pb^p$$ and if $p$ is even: $$(a^2+b^2)^p=a^{2p}+...+\binom {p} {p/2}a^pb^p+...+b^{2p}$$ it is trivial.

If $p$ is odd: $$(a^2+b^2)^p=a^{2p}+...+\binom {p} {(p-1)/2}a^{p-1}b^{p+1}+\binom {p} {(p+1)/2}a^{p+1}b^{p-1}+...+b^{2p}$$ $$\Rightarrow\binom {p} {(p-1)/2}a^{p-1}b^{p-1}(a^2+b^2)\ge 2\binom {p} {(p-1)/2}a^{p-1}b^{p-1}ab\ge 2a^pb^p$$

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I will give another approach. Note that by homogenity, it suffices to check the inequality is true for all $t\geq 0$: $$ (1+t^2)^{\frac{p}{2}} -t^p -1 \geq 0.$$

Set $\phi(t)=(1+t^2)^{\frac{p}{2}} -t^p -1$. Then note $\phi(0)=0$. Then by taking derivative, we have \begin{align*} \phi^\prime (t)&=\frac{p}{2} (1+t^2)^{\frac{p}{2} -1}(2t) -p t^{p-1}\\ &=p t\left((1+t^2)^{\frac{p-2}{2}}-t^{p-2}\right). \end{align*} Hence $\phi^\prime (t)\geq 0$ for all $t\geq 0$. This implies the inequality is true.

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