4
$\begingroup$

I have recently started learning proofs involving inequalities and came across the AM-GM inequality, which seems like a quite powerful tool.

However, I am not sure I understand how to use this tool properly, and I was wondering if there are acceptable strategies to be aware of when making use of the AM-GM inequality.


I have also tried solving an inequality involving AM-GM to learn how to use this tool as I go, but I'm not sure if what I've done so far is a valid approach.

Here is the question:

Prove that if $x, y, z, w ≥ 0$,

$\frac{x+y+z+w}{4} ≥ \sqrt[4]{xyzw}$

Here is what I have done so far:

I noticed that $\sqrt[4]{xyzw}$ = $\sqrt{\sqrt{xy}\sqrt{zw}}$, and then used AM-GM, like so:
I let
$a = \sqrt{xy}$
$ b= \sqrt{zw}$

to make a use of $\frac{a+b}{2} ≥ \sqrt{ab}$.
And, then, I subbed in the values:

$\frac{\sqrt{xy}+\sqrt{zw}}{2} ≥ \sqrt{\sqrt{xy}\sqrt{zw}}$.

Now I have a feeling I'm supposed to somehow make a use of the inequality ($\frac{a+b}{2} ≥ \sqrt{ab}$) once more, however, I am not exactly sure how I should go about doing it.

Any tips for a newbie like myself would be immensely helpful! (Perhaps a link I can refer to, an article, etc)

Thanks a lot in advance for any help!

$\endgroup$
5
  • $\begingroup$ What's wrong with using AM-GM on four numbers directly? $\endgroup$
    – user239203
    Jan 4 '20 at 23:25
  • $\begingroup$ I just picked this inequality as a starting point to prove it using AM-GM for two variables first, and get exposed to the AM-GM inequality. I am eager to learn how to use this tool when dealing with inequalities requiring formal mathematical proofs. $\endgroup$ Jan 4 '20 at 23:28
  • $\begingroup$ Did you mean $\sqrt{xy}+\sqrt{zw}$ where you wrote $\sqrt{zw}+\sqrt{zw}$? $\endgroup$ Jan 4 '20 at 23:34
  • $\begingroup$ Yes, I'm sorry! Fixed. $\endgroup$ Jan 4 '20 at 23:36
  • 1
    $\begingroup$ This doesn't specifically pertain to the problem, but in general here are two sources that may help you: artofproblemsolving.com/wiki/index.php/… brilliant.org/wiki/arithmetic-mean-geometric-mean $\endgroup$
    – Zhuli
    Jan 4 '20 at 23:51
6
$\begingroup$

Start with $\dfrac{a+b}{2} \ge \sqrt{ab} $. To prove this, write it as $\dfrac{a-2\sqrt{ab}+b}{2} \ge 0 $, and the left side is $\dfrac{(\sqrt{a}-\sqrt{b})^2}{2} \ge 0 $.

Then,

$\begin{array}\\ \dfrac{a+b+c+d}{4} &=\dfrac{a+b}{4}+\dfrac{c+d}{4}\\ &=\dfrac{\dfrac{a+b}{2}}{2}+\dfrac{\dfrac{c+d}{2}}{2}\\ &\ge\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{cd}}{2}\\ &=\dfrac{\sqrt{ab}+\sqrt{cd}}{2}\\ &\ge\sqrt{\sqrt{ab}\sqrt{cd}}\\ &=\sqrt{\sqrt{abcd}}\\ &=\sqrt[4]{abcd}\\ \end{array} $

By induction on $n$, with this technique you can show that $\dfrac{\sum_{k=1}^{2^n}a_k}{2^n} \ge \sqrt[2^n]{\prod_{k=1}^n a_k} $.

To show this is true for any $m < 2^n$, let $a_j =\dfrac{\sum_{k=1}^m a_k}{m} $ for $j \gt m$ and see what happens.

As a matter of fact, this was Cauchy's original proof.

Here's the details (added later).

The left side is, letting $a = \dfrac{\sum_{k=1}^m a_k}{m} $,

$\begin{array}\\ \dfrac{\sum_{k=1}^{2^n}a_k}{2^n} &=\dfrac{\sum_{k=1}^{m}a_k}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a_k}{2^n}\\ &=\dfrac{\sum_{k=1}^{m}a_k}{m}\dfrac{m}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a}{2^n}\\ &=\dfrac{am}{2^n}+\dfrac{(2^n-m)a}{2^n}\\ &=\dfrac{am}{2^n}+\dfrac{2^na}{2^n}-\dfrac{ma}{2^n}\\ &= a\\ &=\dfrac{\sum_{j=1}^ma_j}{m}\\ \end{array} $

Similarly, the right side is, letting $a_j =b =\left(\prod_{k=1}^{m} a_k\right)^{1/m} $ for $j > m$,

$\begin{array}\\ \sqrt[2^n]{\prod_{k=1}^{2^n} a_k} &=\left(\prod_{k=1}^{2^n} a_k\right)^{1/2^n}\\ &=\left(\prod_{k=1}^{m} a_k\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\ &=\left(\prod_{k=1}^{m} a_k\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\ &=\left(b^m\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} b\right)^{1/2^n}\\ &=b^{m/2^n}\left(b^{2^n-m}\right)^{1/2^n}\\ &=b^{m/2^n}b^{(2^n-m)/2^n}\\ &=b\\ &=\left(\prod_{k=1}^{m} a_k\right)^{1/m}\\ \end{array} $

Therefore $a \ge b$ or $\dfrac{\sum_{j=1}^ma_j}{m} \ge \left(\prod_{k=1}^{m} a_k\right)^{1/m} $.

$\endgroup$
1
$\begingroup$

$$\frac{x+y+z+w}{4}\ge\frac{\sqrt{xy}+\sqrt{zw}}{2}\ge\sqrt[4]{xyzw}$$

$\endgroup$
6
  • 1
    $\begingroup$ Thanks a lot! Could you just explain in further detail what you did here? I am quite new to proofs and more specifically inequality proofs involving AM-GM. $\endgroup$ Jan 4 '20 at 23:31
  • 1
    $\begingroup$ @FlavioEsposito I used the $2$-variables AM-GM three times, twice to create two square roots & once more for the fourth root. $\endgroup$
    – J.G.
    Jan 4 '20 at 23:40
  • $\begingroup$ I see. So, just to make sure I got it right: You used AM-GM once for $\sqrt{xy}$, once for $\sqrt{zw}$, and then for $\sqrt[4]{xyzw}$? $\endgroup$ Jan 4 '20 at 23:47
  • 1
    $\begingroup$ Yep, that's right! $\endgroup$
    – Zhuli
    Jan 4 '20 at 23:50
  • 1
    $\begingroup$ For any values $a$, $b$, $x$, and $y$, if $\color{red}{a}\ge \color{red}{x}$ and $\color{blue}{b} \ge \color{blue}{y}$, then $\color{red}{a}+\color{blue}{b} \ge \color{red}{x}+\color{blue}{y}$. To see why: $$ a \ge x \\ a+b \ge x+b \ge x+y $$ We know that $\color{red}{\frac{x+y}{2}} \ge \color{red}{\sqrt{xy}}$ and $\color{blue}{\frac{z+w}{2}} \ge \color{blue}{\sqrt{zw}}$ by AM-GM. Apply the result above and you have that relation. $\endgroup$
    – Zhuli
    Jan 6 '20 at 15:08

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.