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!

  • $\begingroup$ 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

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 $.


$\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} $.



  • 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

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