Can every inequality be proved using the AM-GM inequality?

Question

I recently started learning about the AM-GM inequality and wanted to understand how to utilize it to prove different inequalities, and I was questioning if every inequality can be proved using that method?

I was also wondering how one can find the values of $x$ and $y$ (that are later substitued into $\frac{x+y}{2}≥\sqrt{x∗y}$ to prove the given inequality) to make use of the AM-GM? As an example, in a question that I have asked before:

Prove that if $b≥-1$, $b≠0$, then $\frac{4b^2+b+1}{4|b|} ≥ \sqrt{b+1}$

A kind enough user, @Michael Rozenberg gave me an answer of how this can be solved using the AM-GM, however, I wasn't sure how did he know to let $x=|b|$, and $y=\frac{b+1}{4|b|}$, where $\frac{x+y}{2}≥\sqrt{x∗y}$ in order to nicely get the answer:$$\frac{4b^2+b+1}{4|b|}=|b|+\frac{b+1}{4|b|}\geq2\sqrt{|b|\cdot\frac{b+1}{4|b|}}=\sqrt{b+1}$$

Also, I am not quite sure how the AM-GM can applied to this question?

I managed to prove it without using AM-GM like so:
$$\frac{4}{(\frac{1}{a}+\frac{1}{b})^2} \leq ab$$ $$ab(\frac{a+b}{ab})^2-4 \ge 0$$ $$\frac{a^2+2ab+b^2-4ab}{ab} \ge 0$$ $$\frac{(a-b)^2}{ab} \ge 0$$However, I am not sure how to use AM-GM to prove it, and how to find the neccessary values of $x$ and $y$ to do so.

I apologize for the long post.
Any bit of help would be immensely appreciated!

Answer 1

As mentioned in the comments, AM-GM is merely one result, and manipulating an inequality into a certain form required to use AM-GM is just one technique. There's no reason to expect that this is technique will be universal and can enable us to prove any inequality.

With that said, as Robert Wolfe hinted in the comments your particular inequality does have a solution using AM-GM. Taking the reciprocal of both sides and multiplying throughout by $2$, we get $$\frac1a+\frac1b\geq\frac{2}{\sqrt{ab}}$$ This is, of course, equivalent to AM-GM with $x=1/a,y=1/b$ so that $x+y\geq2\sqrt{xy}$ (note that we've implicitly used the fact that $a,b>0$).

Answer 2

The left hand side is the harmonic mean of $a$ and $b$, which is less than or equal to the geometric mean. If you wanted to use AM-GM, note that you may construct a new circle whose radius is the geometric mean and altitude is the harmonic mean.

Answer 3

Also note that AM-GM for 2 variables can be proved without squaring by writing $\dfrac{x+y}{2} \ge \sqrt{xy} $ as $\dfrac{x+y-2\sqrt{xy}}{2} \ge 0 $ or $\dfrac{(\sqrt{x}-\sqrt{y})^2}{2} \ge 0 $.

Similarly, $\dfrac1a+\dfrac1b\geq\dfrac{2}{\sqrt{ab}} $ is the same as $(\dfrac1{\sqrt{a}}-\dfrac1{\sqrt{b}})^2 \ge 0 $. No squaring or reciprocating.