# how to prove the inequality $\sqrt[n]{\frac{a^n+b^n}{2}}\geq\frac{a+b}{2}$?

given natural number $n$, how to prove the inequality $\sqrt[n]{\frac{a^n+b^n}{2}}\geq\frac{a+b}{2}$?

furthermore, given real number x, how to prove $f(x)=\sqrt[x]{\frac{a^x+b^x}{2}}$ increasing?

• Well. try proving $2^n(a^n + b^n)\ge 2(a+b)^n$. – fleablood Oct 10 '17 at 1:45
• thats a good idea. but how about proving its increasing. – Charles Bao Oct 10 '17 at 1:50
• For which $n$? For which $x$? – robjohn Oct 10 '17 at 2:52
• @robjohn for natural number $n$, real number $x$. – Charles Bao Oct 10 '17 at 3:03

Let $f(x)=x^n$.

Thus, $f$ is a convex function.

Thus, by Jensen we obtain: $$\frac{f(a)+f(b)}{2}\geq f\left(\frac{a+b}{2}\right)$$ or $$\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$$ or $$\sqrt[n]{\frac{a^n+b^n}{2}}\geq\frac{a+b}{2}.$$

Now, let $x\geq y>0$.

We'll prove that $$\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}\geq\left(\frac{a^y+b^y}{2}\right)^{\frac{1}{y}}.$$ Indeed, let $a^y=p$, $b^y=q$ and $\frac{x}{y}=k$.

Thus, $k\geq1$ and we need to prove that $$\frac{p^k+q^k}{2}\geq\left(\frac{p+q}{2}\right)^k,$$ which is Jensen again.

Done!

Use strong induction with $$a^{n+1}+b^{n+1}=(a+b)(a^{n}+b^{n})-ab(a^{n-1}+b^{n-1}).$$

Using Minkowski's inequality superadditivity of the geometric mean, one has $$\sqrt[n]{x^n}+\sqrt[n]{y^n} \leq \sqrt[n]{2^{n-1}(x^n+y^n)}$$

$$(x+y) \leq \sqrt[n]{2^{n-1}(x^n+y^n)}$$

$$\frac{x+y}{2} \leq \sqrt[n]{\frac{x^n+y^n}{2}}.$$

If $x\ge y\gt0$, then $u^{x/y}$ is a convex function of $u$. Therefore, Jensen's inequality says that $$\frac{a^{x/y}+b^{x/y}}2\ge\left(\frac{a+b}2\right)^{x/y}$$ Then substituting $a\mapsto a^y$ and $b\mapsto b^y$ gives $$f(x)=\left(\frac{a^x+b^x}2\right)^{1/x}\ge\left(\frac{a^y+b^y}2\right)^{1/y}=f(y)$$ This shows that $f(x)$ is increasing. Furthermore, setting $x=n\ge1$ and $y=1$ gives the first inequality.

Raise both sides by $n$, rearrange.

$$\underbrace{(1^n+1^n)(1^n+1^n)\cdots (1^n+1^n)}_{n-1\text{ times}}(a^n+b^n)\ge$$

$$\ge (a+b)^n$$

This is true by Hölder’s inequality.