Let $0 < a \le b$. Then $a^2 \le \left (\frac{2ab}{a+b} \right) ^2 \le ab \le \left( \frac{a+b}{2} \right) ^2 \le b^2$

Question

In a introductory text about real analysis, in a chapter about the axioms of order (I hope this is the correct translation), I was faced with the following

Problem: Let $0 < a \le b$. Show that $a^2 \le \left (\frac{2ab}{a+b} \right) ^2 \le ab \le \left( \frac{a+b}{2} \right) ^2 \le b^2$. If, at any point of this chain of inequalities, we have equality, then $a = b$.

There was no solution given. I came to the following

Solution: The first inequality follows from

$$ \begin{align} a^2 & \le ab\\ a^2 + ab & \le 2ab \\ a(a+b) & \le 2ab \\ a & \le \frac{2ab}{a+b} \\ a^2 & \le \left( \frac{2ab}{a+b} \right) ^2 \end{align}$$

By the same reasoning, if equality holds, $a^2=ab$ implies that $a=b$.

We get the second inequality from

$$ \begin{align} 0 & \le (a-b)^2\\ 0 & \le a^2 - 2ab + b^2 \\ 4ab & \le a^2 + 2ab + b^2 \\ \frac{4a^2b^2}{ab} & \le (a+b)^2 \\ \frac{4a^2b^2}{(a+b)^2} & \le ab \\ \left( \frac{2ab}{a+b} \right) ^2 & \le ab \end{align}$$

an similarly for the third inequality with

$$ \begin{align} 0 & \le (a-b)^2\\ 0 & \le a^2 - 2ab + b^2 \\ 4ab & \le a^2 + 2ab + b^2 \\ ab & \le \frac{(a+b)^2}{4} \\ ab & \le \left( \frac{(a+b)}{2} \right) \\ \end{align}$$

In both cases, $0 = (a-b)^2$ implies that $a=b$.

Finally we have

$$ \begin{align} a & \le b \\ a + b & \le 2b \\ \frac{a+b}{2} & \le b \\ \left( \frac{a+b}{2} \right) ^2 & \le b^2 \end{align}$$

and equality directly leads to $a=b$. $\blacksquare$

I have three questions to this:

1. Is my solution correct?
2. Do I have to take additional steps for the second part of the problem, that equality implies $a=b$?
3. Are there shorter and more elegant solutions to this?

Thanks in advance!

Answer 1

It look quite correct for me, and the equality part is good since $a,b$ are both positive.

Just for showing a shorter (maybe) version for this by using two important but simple inequalies:

(1). $(a+b)^2 \geq 4ab$

And its variation in case of $a>0,b>0$ :

(2). $ 1 \geq \frac{4ab}{(a+b)^2}$

We have: $$ b^2 \geq (\frac{a+b}{2})^2 \ \ (\text{because } b \geq a > 0) \\ = \frac{(a+b)^2}{4} \geq \frac{4ab}{4}=ab \ \ \ (\text{because (1) }) \\ = ab \times 1 \geq ab \times \frac{4ab}{(a+b)^2} = (\frac{2ab}{a+b})^2 \ \ \ (\text{because (2) }) \\ \geq (\frac{2ab}{b+b})^2 = a^2 \ \ \ (\text{because } b \geq a >0 ) $$

Actually, the first and the last inequalites use $a\geq b>0$, the second and third ones use HM-GM-AM-QM inequalities mentioned by @mrtechtroid.