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Let $a,b\in[0,\infty)$ and let $p\in[1,\infty)$. How can I prove$$a^p+b^p\le(a^2+b^2)^{p/2}.$$

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I think you should check and change the inequality sign, as when a=b=p=1 the inequality is not satisfied. – Tomarinator May 27 '12 at 9:52
To expand on the comment by @DavideGiraudo, take the $p$'th root and consider $(a^p+b^p)^{1/p}$ as a function of $p$. – Harald Hanche-Olsen May 27 '12 at 10:22

Some hints:

  • By homogeneity, we can assume that $b=1$.
  • Let $f(t):=(t^2+1)^{p/2}-t^p-1$ for $t\geq 0$. We have $f'(t)=p((t^2+1)^{p/2-1}-t^{p-1})$. We have $t^2+1\geq t^2$, so the derivative is non-negative/non-positive if $p\geq 2$ or $p<2$.
  • Deduce the wanted inequality (when it is reversed or not).
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