# Proving tricky inequality

For positive a,b prove that : $\frac{2}{(1/a)+(1/b)}≤\sqrt{ab}≤\frac{(a+b)}{2}$

I think the solution is multiply through to get

$4ab≤2(a+b)\sqrt{ab}≤(a+b)^2$

From this we can deduce that if:

$(a+b)≥2\sqrt{ab}$

then the inequality holds and you can prove this by starting with:

$(a-b)^2 ≥0$ and then rearrange to $(a+b)^2 ≥4ab$ and then take the square root?

This seems too easy so I am not sure.

Any help much appreciated.

• no because if you look at the one in the centre and the one on the left. Divide both by $2\sqrt{ab}$ and you get my 3rd equation above. Do the same for centre and equation on the right and you get the same inequality so to prove each one separately you just have to prove the same thing..? – Cam Mack Oct 29 '15 at 23:43
• Each can be obtained, in slightly different ways, from the inequality $(\sqrt{a}-\sqrt{b})^2\ge 0$. – André Nicolas Oct 29 '15 at 23:46
• So is my method wrong or just not the best way? – Cam Mack Oct 29 '15 at 23:48
• Oh I see, yours is similar but you just added square roots – Cam Mack Oct 29 '15 at 23:50
• I take it back. Your calculation is fine. – André Nicolas Oct 29 '15 at 23:55

We first prove the inequality of arithmetic and geometric means $$\sqrt{ab} \leqslant \frac{a+b}{2}.$$ It results from $$\frac{a+b}{2} - \sqrt{ab} = \frac{(\sqrt{\vphantom{b}a})^2 - 2\sqrt{\vphantom{b}a}\sqrt{b} + (\sqrt{b})^2}{2} = \frac{(\sqrt{\vphantom{b}a}-\sqrt{b})^2}{2} \geqslant 0.$$ The inequality of geometric and harmonic means $$\frac{2}{1/a + 1/b} \leqslant \sqrt{ab}$$ can be easily obtained by applying the AM-GM inequality to the numbers $1/a$ and $1/b$: $$\sqrt{\frac{1}{ab}}\leqslant \frac{1/a + 1/b}{2},$$ then taking the inverse.