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:
- Is my solution correct?
- Do I have to take additional steps for the second part of the problem, that equality implies $a=b$?
- Are there shorter and more elegant solutions to this?
Thanks in advance!