# Analog of $(a+b)^2 \leq 2(a^2 + b^2)$

Is there an inequality such as $$(a+b)^2 \leq 2(a^2 + b^2)$$ for higher powers of $k$ $$(a+b)^k \leq C(a^k + b^k)?$$

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The generalized mean inequality states that $$\dfrac{a+b}2\leq \left(\dfrac{a^k+b^k}2\right)^{1/k},$$ with equality if and only if $a=b$, from which it follows that $$(a+b)^k\leq 2^{k-1}(a^k+b^k),$$ with equality if and only if $a=b$. Thus $C=2^{k-1}$ works and no smaller $C$ works.

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Take the ratio $$f(a, b)=\frac{(a+b)^k}{a^k+b^k}$$and observe that it is homogeneous of degree zero, that is $f(a, b)=f(\lambda a, \lambda b)$ for all $\lambda >0$. So $f$ is constant along rays in $\mathbb{R}^2$ and so, in particular, $$f(a, b)\le \max_{(x, y)\in \mathbb{S}^1} f(x, y),$$ where $\mathbb{S}^1$ is the unit circle. This means that the sought inequality is true with $C=\max_{\mathbb{S}^1} f(x,y)$. You can check that with $k=2$ you recover $C=2$.

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Too pretty man. +1 –  user63477 Mar 26 at 12:24
The first inequality is true because : $2ab ≤ a^2+b^2$
Yes. I have tried with $C=2^k$ and it seems it doesn`t work. –  user67878 Mar 26 at 10:40
I just noticed that $C = 2^{k-1}$ works (math.stackexchange.com/questions/215472/…). –  hannah Mar 26 at 10:42
It seems that $c=2^k$ should work as Samuel stated it above, my intuition seems to be good but it looks that I have failed with calculations. –  user67878 Mar 26 at 10:42