Proving the Inequality |x+y|^p \leq 2^p(|x|^p+|y|^p)

[quasar987]

[SOLVED] Help me prove this inequality

Homework Statement

The inequality in question is

$$|x+y|^p \leq 2^p(|x|^p+|y|^p)$$

for any positive integer p and real numbers x,y.

The Attempt at a Solution

For p=1, it is weaker than the triangle inequality.

Suppose it is true for p, and let's try to show this implies it's true for p+1.

$$|x+y|^{p+1}=|x+y||x+y|^p\leq |x+y|2^p(|x|^p+|y|^p)$$

And basically, here I've tried using the triangle inequality on |x+y| but the most "reduced form" I got is I arrived at the conclusion that the inquality was true iff

$$|x||y|(|x|^p+|y|^p)\leq |x|^{p+1}+|y|^{p+1}$$

 

[Kummer]

Let $$a,b\geq 0$$ we want to show $$(a+b)^n \leq 2^n(a^n+b^n)$$. First accept induction. Then $$(a+b)^{n+1} = (a+b)^n(a+b)\leq 2^n(a^n+b^n)(a+b) = 2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)$$. But $$2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)\leq 2^{n+1}(a^{n+1}+b^{n+1})$$ iff $$a^{n+1}+b^{n+1}\leq 2(a^{n+1}+b^{n+1}+a^nb+ab^n)$$ iff $$a^{n+1}+b^{n+1}\geq ab^n+a^nb$$ iff $$(a-b)(a^n-b^n)\geq 0$$ but that is true because $$(a-b)(a^n-b^n) = (a-b)^2(a^{n-1}b+...+ab^{n-1})\geq 0$$.

 

[quasar987]

But $$2^n(a^{n+1}+b^{n+1}+ab^n+a^nb)\leq 2^{n+1}(a^{n+1}+b^{n+1})$$ iff $$a^{n+1}+b^{n+1}\leq 2(a^{n+1}+b^{n+1}+a^nb+ab^n)$$

Ok, this is just a typo probably because you fall back on your feet a few lines later with

$$a^{n+1}+b^{n+1}\geq ab^n+a^nb$$

Good work, thanks Kummer.

 

[quasar987]

How do I add [SOLVED] to the title?

 

[Kummer]

How do I add [SOLVED] to the title?

Click on "thread tools".

 

[Kurdt]

As a homework helper quasar you can do that for any thread.

https://www.physicsforums.com/showthread.php?t=184283