How is AM-GM supposed to be used to prove an inequality

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!

• What's wrong with using AM-GM on four numbers directly?
– user239203
Jan 4 '20 at 23:25
• 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. Jan 4 '20 at 23:28
• Did you mean $\sqrt{xy}+\sqrt{zw}$ where you wrote $\sqrt{zw}+\sqrt{zw}$? Jan 4 '20 at 23:34
• Yes, I'm sorry! Fixed. Jan 4 '20 at 23:36
• 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 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 $$\dfrac{(\sqrt{a}-\sqrt{b})^2}{2} \ge 0$$.

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

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

• 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. Jan 4 '20 at 23:31
• @FlavioEsposito I used the $2$-variables AM-GM three times, twice to create two square roots & once more for the fourth root.
– J.G.
Jan 4 '20 at 23:40
• 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 $\sqrt[4]{xyzw}$? Jan 4 '20 at 23:47
• Yep, that's right! Jan 4 '20 at 23:50
• 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. Jan 6 '20 at 15:08