Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume that you have any two real numbers $a$ and $b$, and $1\leq k <\infty$, $k \in \mathbb{R}$.

How would you prove the inequality $|a+b|^k \leq 2^{k-1} (|a|^k+|b|^k)$?

share|cite|improve this question
What about $k$? Is it a positive integer? A positive real? – Patrick Da Silva Oct 17 '12 at 7:16
Edited, $1 \leq k < \infty$. – Rojas Azules Oct 17 '12 at 7:17
Sorry, $k \in \mathbb{R}$. – Rojas Azules Oct 17 '12 at 7:20
up vote 3 down vote accepted

Method $1$:

The function $f(x) = x^k$ for $x\geq 0$ is a convex function for $k \geq 1$. Now apply Jensen's inequality.

Method $2$:

Let $\lvert a \rvert \geq \lvert b \rvert$. Let $t = \dfrac{b}a \implies 0 \leq \vert t \vert \leq 1$. We then want to prove that $$\vert 1 + t \vert^k \leq 2^{k-1} \left(1+\vert t \vert^k \right)$$ $$\left \vert \dfrac{1+t}2 \right \vert^k \leq \dfrac{1+\left \vert t \right \vert^k}2 $$ $$\left \vert \dfrac{1+t}2 \right \vert^k \leq \left(\dfrac{1+ \vert t \vert}{2} \right)^k$$ Setting $y = \vert t \vert \in [0,1]$, we want to prove that $$\left(\dfrac{1+ y}{2} \right)^k \leq \dfrac{1+y^k}2$$ $$f(y) = \dfrac{1+y^k}2 - \left(\dfrac{1+ y}{2} \right)^k$$ $$f'(y) = \dfrac{ky^{k-1}}2 - \dfrac{k}{2^k}(1+y)^{k-1} = \dfrac{k}2 \left(y^{k-1} - \left(\dfrac{1+y}2 \right)^{k-1} \right)$$ For $y \in [0,1]$, $y \leq \dfrac{1+y}2$, and hence $y^k \leq \left( \dfrac{1+y}2 \right)^k$. This means that $f'(y) < 0$ and hence $f(y)$ is decreasing. Hence, $$f(y) > f(1)$$ which implies $$\left(\dfrac{1+ y}{2} \right)^k \leq \dfrac{1+y^k}2$$

share|cite|improve this answer
I was just about to write exactly this. Haha – Patrick Da Silva Oct 17 '12 at 7:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.