# Prove inequality $|tx+sy|^p\le t|x|^p+(1-t)|y|^p$

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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What does not work for $x\leq 0$? –  Dirk Feb 6 '13 at 9:23
how can you prove that $f$ is covex for $x\le 0$? –  Johnny Feb 6 '13 at 9:30
E.g. check if the derivative is nondecreasing. –  Dirk Feb 6 '13 at 9:33

## 1 Answer

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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The composition of two convex functions is not necessarily convex.. –  Mårten W Feb 6 '13 at 9:56
I have corrected it. –  Gautam Shenoy Feb 6 '13 at 10:02
I have taken back my vote. But in my world, "the first function" in a composition is the innermost.. –  Mårten W Feb 6 '13 at 10:07
Thanks Marten. I have cleared the ambiguity. –  Gautam Shenoy Feb 6 '13 at 10:10