Question

Question: Prove AM-GM-HM inequality. (additional question) Can it find the exact extrema?

Answer 1

Let's start with AM ≥ GM proof.

If we finish the proof of (AM) - (GM) ≥ 0 then it is clear.

First, the AM - GM is

→ $\dfrac{a+b}{2} -\sqrt{ab}$

→ $\dfrac{a-2\sqrt{ab} +b}{2}$

→ $\dfrac{(\sqrt{a} -\sqrt{b} )^2}{2} \geq 0$ Hence proven.

Then we finish with the GM ≥ HM proof.

Like previous let's go for GM - HM ≥ 0.

First, the GM - HM is

→ $\sqrt{ab} -\dfrac{2ab}{a+b}$

→ $\dfrac{\sqrt{ab} (a+b)-2ab}{a+b}$

→ $\dfrac{\sqrt{ab} (a-2\sqrt{ab} +b)}{a+b}$

→ $\dfrac{\sqrt{ab} (\sqrt{a} -\sqrt{b} )^2}{a+b}\geq 0$ Hence proven.

That is to say, AM ≥ GM ≥ HM. (Equality where a=b)

---

Sure.

In a situation without given conditions, the AM-GM-HM can find the exact maxima or minima.

For example, let there be a rectangle which sides are $a$ and $b$.

For a rectangle of the perimeter 40, $2a+2b=40$ and therefore

$\dfrac{a+b}{2} =10$

Which is AM here. So the GM cannot exceed it therefore

$\sqrt{ab} \leq 10$

Furthermore obtaining

Area: $ab\leq 100$

Answer 2

8888888888888888888888