# Is it possible to prove that $|a+b|^p \leq |a|^p+|b|^p$?

I'm trying to prove that the function $f:\mathbb{R}\to \mathbb{R}$, $$f(x) = |x|^p \quad ,\quad p \geq 1$$ is convex. By using the definition of a convex function and simplifying a bit, I arrived in the following inequality $$|a+b|^p \leq |a|^p+|b|^p$$ where a and b are real numbers.

If I can prove that this holds, then $f$ is convex. However, I'm a bit lost here. I'm aware of the triangle inequality, but that only proves it for $p=1$. Is there such thing as a triangle inequality to the power of $p$? If not, any other suggestions would appreciated.

• Let $p = 2$ and $x = y = 0.1$. Then $$|x+y|^p = |0.1 + 0.1|^2 = 0.2^2 = 0.04 > 0.02 = 0.01 + 0.01 = |0.1|^2 + |0.1|^2 = |x|^p + |y|^p$$ – TastyRomeo Nov 9 '16 at 10:24
• Since you write $\lVert x\rVert^p$, I guess the domain of $f$ should be $\mathbb{R}^n$ rather than $\mathbb{R}$? Look at the function $g\colon [0,+\infty) \to [0,+\infty)$ given by $g(t) = t^p$. And think about the composition of convex functions. – Daniel Fischer Nov 9 '16 at 10:24
• @SteamyRoot: thank you for the counter-example. However, it now turned the question moot. What's the best practice here, delete it? – JLagana Nov 9 '16 at 10:30
• @DanielFischer: No, I actually meant $|x|$, instead of $||x||$. Thanks for pointing it out. – JLagana Nov 9 '16 at 10:31
• Since $f$ is continuous, it suffices to prove midpoint convexity to deduce convexity. And that is $$\biggl\lvert \frac{a+b}{2}\biggr\rvert^p \leqslant \frac{\lvert a\rvert^p + \lvert b\rvert^p}{2}.$$ – Daniel Fischer Nov 9 '16 at 10:37