Struggling with this proof.

Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.

What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?

Thanks in advance.

  • $\begingroup$ Have you tried binomial expansion? $\endgroup$ Aug 26, 2015 at 18:25

6 Answers 6


Solution 1.(Partial solution) $x^n$ is a convex function on $\mathbb{R}^+$ for $n\geq 1$, thus by Jensen: $$\left(\frac{a+b}{2}\right)^n\leq \frac{a^n+b^n}{2}$$

Solution 2.('Hidden' usage of condition) For $n=1$, true. By induction: $$\left(\frac{a+b}{2}\right)^n=\left(\frac{a+b}{2}\right)^{n-1}\left(\frac{a+b}{2}\right)\leq \frac{a^{n-1}+b^{n-1}}{2}\cdot\frac{a+b}{2} $$ And: $$\frac{a^n+b^n}{2}-\frac{a^{n-1}+b^{n-1}}{2}\cdot\frac{a+b}{2}=\frac{(a^{n-1}-b^{n-1})(a-b)}{4}$$ And the factors have the same sign.

Solution 3. Let $\alpha=\frac{a+b}{2}$, then $a=\alpha+x$ and $b=\alpha-x$, then the LHS is $\alpha^n$, and the RHS-LHS is: $$ \frac{1}{2}(a^n+b^n)-\left(\frac{a+b}{2}\right)^n=\sum_{i=1}^{\lfloor n/2\rfloor} \binom{n}{2i}\alpha^{n-2i}x^{2i} $$ And by the conditions, all the terms are positive.

  • 1
    $\begingroup$ Solution (1) doesn't clearly handle the case where $n$ is odd and $a<0<b$. Where are you using that $\frac{a+b}{2}>0$? Where in the second half? $\endgroup$ Aug 26, 2015 at 18:47
  • $\begingroup$ In the induction step, by using that it is true for the previous one (at the first inequality). Thank you for pointing out the error of the first solution. $\endgroup$ Aug 26, 2015 at 19:01
  • $\begingroup$ I also vaguely object to using the phrase "Jensen's inequality." This is actually the definition of convex - by asserting convexity, then using "Jensen's," you are basically missing that you are assuming what you are trying to prove. Jensen's inequality is a general thing, which has follows from the definition of convex. $\endgroup$ Aug 26, 2015 at 19:05
  • $\begingroup$ I really don't get the second solution after the "And: " part. Can somebody give me more details on what is happening? $\endgroup$
    – Nikola
    Jun 30, 2017 at 7:23

When $n$ is a positive integer, the function $f(x) = x^n$ is convex on $(0, +\infty)$, thus by Jensen's inequality: $$f\left(\frac{a}{2} + \frac{b}{2}\right) \leq \frac{1}{2}f(a) + \frac{1}{2}f(b),$$ gives the desired inequality.

As pointed out by Andrews, the $f$ defined above is convex on the whole line only when $n$ is even for which case Jensen's inequality can be applied directly. The inequality also holds if both $a$ and $b$ are nonnegative.

To the case that $n$ is an odd positive integer and $a \leq 0 < b$ (without losing of generality), $a + b \geq 0$, write $n = 2k + 1$, then the right hand side of the inequality is $$\frac{a^{2k + 1} + b^{2k + 1}}{2} = \frac{1}{2}(a + b)(a^{2k} - a^{2k - 1}b + \cdots + b^{2k}) \geq \frac{1}{2}(a + b)(a^{2k} + b^{2k}) > \frac{a + b}{2}\frac{a^{2k} + b^{2k}}{2}$$

So if we can show $$\left(\frac{a + b}{2}\right)^{2k} \leq \frac{a^{2k} + b^{2k}}{2}$$ the inequality holds, but this is the case which we can use Jensen's inequality.

  • $\begingroup$ Actually, $f(x)=x^n$ is only convex on the entire real line when $n$ is even. $\endgroup$ Aug 26, 2015 at 18:45
  • $\begingroup$ well done +1 :-) $\endgroup$
    – Math-fun
    Aug 26, 2015 at 18:56
  • $\begingroup$ I'll write what I wrote to another answer: I also vaguely object to using the phrase "Jensen's inequality." This is actually the definition of convex - by asserting convexity, then using "Jensen's," you are basically missing that you are assuming what you are trying to prove - that $x^n$ is convex. $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$ is the definition of convex, not a result of convexity. $\endgroup$ Aug 26, 2015 at 19:32

A much easier (than my original) answer

Let $c>0$ and let $$p_c(x)=(c+x)^n +(c-x)^n=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}c^{n-2k}x^{2k}$$ Note that $p_c(x)$ has only positive coefficients and every term has even power. So $p_c(0)\leq p_c(x)$ for all $x$.

Then let $c=\frac{a+b}{2}$ and $x=\frac{a-b}{2}$. Then $c+x=a$ and $c-x=b$, and we get:

$$a^n + b^n =p_c(x)\geq p_c(0) = c^n + c^n=2\left(\frac{a+b}{2}\right)^n$$

My original answer


Let $$f(x,y)=\sum_{i=0}^{n-1} x^{i}y^{n-1-i}.$$ Then we can easily show that $$f(x,y)=f(y,x)\tag{1}$$ $$(x-y)f(x,y)=x^n-y^n\tag{2}$$ $$f(x,y_1)\leq f(x,y_2)\text{ when } x\geq 0\text{ and } |y_1|\leq y_2\tag{3}$$


Let $a\leq b$ and let $c=\frac{a+b}{2}$. Note $b-c=c-a\geq0.$

Now, since $a\leq b$ and $-a< b$, we have $|a|\leq b$ so we have, by $(1)$ and $(3)$:

$$f(c,a)\leq f(c,b)=f(b,c)$$

Multiplying this inequality by $c-a=b-c\geq 0$, we get, by $(2)$:

$$c^n-a^n\leq b^n-c^n.$$

Hence $$\frac{a^n+b^n}{2} \geq c^n,$$ which is the inequality we want.

Sadly, this doesn't work for non-integers $n\geq 1$.


The cases $n=1,2$ are trivial, so lets suppose $n\ge 3$. The inequality is equivalent to $$\left(\frac{2a}{a+b}\right)^n+\left(\frac{2b}{a+b}\right)^n\ge 2$$ Then, it will be sufficient to prove that $f:(0,2)\to \mathbb{R}$ defined by $f(x)=x^n+(2-x)^n$ has a minimum value of $2$. Taking the derivative of $f$ we have $$f'(x)=nx^{n-1}-n(2-x)^{n-1}$$ So, by setting $f'(x)=0$, we find that $f$ has a critical point at $x=1$, since \begin{align*} f''(x)&=n(n-1)x^{n-2}+n(n-1)(2-x)^{n-2}\\ f''(1)&=2n(n-1)\\ f''(1)&>0 \end{align*} Then $f(1)=1^n+1^n=2$ is the minimum value of $f$. By putting $x=\frac{2a}{a+b}$ the inequality follows.


$$\left(\frac{a+b}{2}\right)^n = \frac{ \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k}{2^n}= \frac{ \sum_{k=0}^n \binom{n}{k} a^{k}b^{n-k}}{2^n}$$

Therefore $$\left(\frac{a+b}{2}\right)^n =\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n-k}b^k+a^kb^{n-k})}{2^n}$$

Now for each $k$ we have by AM-GM: $$a^{n-k}b^k \leq\frac{a^{n}+...+a^{n}+b^n+..+b^n}{n}=\frac{(n-k)a^{n}+kb^n}{n} $$ and similarly $$a^{k}b^{n-k} \leq\frac{a^{n}+...+a^{n}+b^n+..+b^n}{n}=\frac{ka^{n}+(n-k)b^n}{n} $$

Therefore, by adding them together we get $$a^{n-k}b^k+a^kb^{n-k}\leq a^n+b^n$$

This yields Therefore $$\left(\frac{a+b}{2}\right)^n =\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n-k}b^k+a^kb^{n-k})}{2^n} \leq\frac{1}{2} \frac{ \sum_{k=0}^n \binom{n}{k} (a^{n}+b^{n})}{2^n}\\=\frac{a^n+b^n}{2} \frac{ \sum_{k=0}^n \binom{n}{k} }{2^n} =\frac{a^n+b^n}{2} $$


Another way to prove. Let $$ f(x)=\left(\frac{a+x}2\right)^n-\frac{1}{2}(a^n+x^n), x\ge a. $$ Then $$ f'(x)=\frac{n}{2^n}(a+x)^{n-1}-\frac{n}{2}x^{n-1}=\frac{n}{2^n}[(a+x)^{n-1}-(2x)^{n-1}]\le0 $$ and hence $f(x)$ is decreasing for $x\ge a$. Then if $a<b$, then $f(b)\le f(a)$, namely, $$ \left(\frac{a+b}2\right)^n\le\frac{1}{2}(a^n+b^n). $$

  • $\begingroup$ What happens if $a<0$? Then, you need to restrict to $x\geq -a$. Or, I suppose, in one step, $x\geq |a|$. $\endgroup$ Aug 26, 2015 at 21:36

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