Let $a,b\in[0,\infty)$ and let $p\in[1,\infty)$. How can I prove$$a^p+b^p\le(a^2+b^2)^{p/2}.$$
How to prove an L$^p$ type inequality
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real-analysis
inequality
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2I think you should check and change the inequality sign, as when a=b=p=1 the inequality is not satisfied. – 2012-05-27
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1To expand on the comment by @DavideGiraudo, take the $p$'th root and consider $(a^p+b^p)^{1/p}$ as a function of $p$. – 2012-05-27