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Prove inequality: for any $x,y,p,s\in\mathbb{R}$ with $p\ge 1$, $t \in [0,1]$, $s+t=1$: $$|tx+sy|^p\le t|x|^p+(1-t)|y|^p$$ Attempts: I tried to define a function $f=|x|^p$ and tried to show that it is a convex function for $x>0$ but seem not work with $x\le 0$ |
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The trick is to show that $x^p$ is convex for $x\geq0$. Then when you replace $x$ with $|x|$, use the fact that composition of two convex functions(say f,g) with f nondecreasing is convex ($x^p$ is non decreasing for positive reals) . You dont have to worry about $x<0$ as the range of $|x|$ is the non-negative reals. To prove $x^p$ is convex, for $p\geq2$, you can use the second derivative test. For p=1, it is linear. For $p \in (1,2)$, the first derivative yields $px^{p-1}$, which is monotonically increasing and hence, convex. |
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