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!
Best Answer
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} $.