# Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n})$

Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n})$ I tried induction theorem, when $n = 1$ it is obviously right. But, say $n=k$, It does not make sense since I cannot expand the $2^{k-1}$($a^{k}$+$b^{k}$).

And I also looked through this similar question but no help.

Prove: $(a + b)^{n} \geq a^{n} + b^{n}$

• An identical question was asked earlier. – Anant Mar 19 '14 at 5:38

## 4 Answers

The simpliest proof:

as $x\to x^n$ is convex, $$\left(\frac {x+y}2\right)^n \le \frac {x^n+y^n}2$$ Now multiply by $2^n$ and you are done.

• Can you explain a little more? Our course do not cover things like the "convex". Do you mean increasing? – SonicFancy Mar 5 '14 at 23:40
• convex mean, in a context of a real valued differential function, mean that the derivative function is increasing. If this is the case, a theorem yields $a_i > 0 \ \&\ \sum a_i = 1 \Rightarrow f(\sum a_i x_i) \le \sum a_i f(x_i)$. For exemple, the absolute value is convex. – mookid Mar 5 '14 at 23:46
• @SonicFancy: why did you unaccept? – mookid Oct 26 '14 at 22:10

First of all, this is only true for positive $a$, $b$.

For the induction step: suppose $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ for all $a,b>0$. Multiplying both sides of the inductive hypothesis by $a+b$ (here we use $a+b>0$), we find $(a+b)^{n+1} \leq 2^{n}(a^{n} + b^{n})(a+b)$. It remains to show that $(a^n+b^n)(a+b) \leq 2(a^{n+1}+b^{n+1})$. This is equivalent to $(a^n-b^n)(a-b) \geq 0$, which is true as $x \mapsto x^n$ is an increasing function.

This is the generalized mean inequality for two variables (comparing the arithmetic mean and the $n$-th power mean).

• After multiplying a+b to the supposed part, is that should still be $2^{n-1}$? – SonicFancy Mar 5 '14 at 23:44
• And should I consider that the basecase is n=1? – SonicFancy Mar 5 '14 at 23:49

$$(a+b)^n\leq 2^{n-1}(a^n+b^n)\Longleftrightarrow\left(\frac{a+b}2\right)^n\leq\frac{a^n+b^n}2\Longleftrightarrow\frac{a+b}2\leq\sqrt[n]\frac{a^n+b^n}2$$ Which is a special case of the generalized mean inequality if $a,b\geq0$ and $n\geq1$ (can even be real).

The full inequality gives you a nice generalized bound $$\left(\sum_{i=1}^k a_i\right)^n\leq k^{n-1}\sum_{i=1}^k a_i^n$$

• This is a little bit hard for me to understand but still thanks a lot. – SonicFancy Mar 5 '14 at 23:46

Also, the TL method helps.

Indeed, since our inequality is homogeneous, we can assume that $$a+b=2$$ and we need to prove that $$a^n+b^n\geq2$$ or $$(a^n-na+n-1)+(b^n-nb+n-1)\geq0,$$ which is true by AM-GM: $$\sum_{cyc}(a^n-nb+n-1)\geq\sum_{cyc}\left(n\sqrt[n]{a^n\cdot1^{n-1}}-na\right)=0$$ and we are done!