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Possible Duplicate:
Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$

We can use Young's inequality to show that $(a+b)^2 \leq 2a^2 + 2b^2$.

Does a similar result hold for the n-th power as well? That is, do we have $(a + b)^n \leq c_1 a^n + c_2 b^n$ ?

If so, what are the values for $c_1$ and $c_2$?

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Symmetry demands that $c_1 = c_2$. $a=b$ gives you the desired values of $c_1,c_2$ namely, $c_1 = c_2 = 2^{n-1}$. You might want to throw in the $\lvert \cdot \rvert$ into the equation i.e. $\lvert a+ b \rvert^{n} \leq 2^{n-1} \lvert a \rvert^n + 2^{n-1} \lvert b \rvert^n$ A very similar question and the proof for the above can be found here.

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    @mkolar http://math.stackexchange.com/questions/143173/showing-the-inequality-alpha-betap-leq-2p-1-alphap-betap2012-05-15