3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Have you tried binomial expansion? $\endgroup$ Aug 26, 2015 at 18:25

6 Answers 6

9
$\begingroup$

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.

$\endgroup$
4
  • 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
5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
3
$\begingroup$

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

Preliminary

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

Proof

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 $n\geq 1$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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, $$ \left(\frac{a+b}2\right)^n\le\frac{1}{2}(a^n+b^n). $$

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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