I'm trying to prove an inequality from an old book named "Algebra and elementary functions" by Kochetkov (ru - "Алгебра и элементарные функции" Е. С. Кочетков). This is an old book from 1970 (URSS era). The book is not copyrighted anymore, and it is freely available over the internet.
The inequality is $\frac{a}{\sqrt{b}} + \frac{b}{\sqrt{a}} \ge \sqrt{a} + \sqrt{b}$ (complex numbers are not allowed)
Here are my steps :
$\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} \ge \sqrt{a} + \sqrt{b} $
$\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} - \sqrt{a}-\sqrt{b} \ge 0$
$\frac{a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a}))}{\sqrt{ab}} \ge 0$
It is clear that if $a=b$, then the expression $a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})$ is zero. Also intuitively I can see that in other cases the expression $a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})$ is greater than 0, but how to prove it formally ?
You have almost solved it. You need to prove that
$a(\sqrt{a}-\sqrt{b})+b(\sqrt{b}-\sqrt{a})\ge 0$.
which is true due to the following :
$a(\sqrt{a}-\sqrt{b})-b(\sqrt{a}-\sqrt{b})\ge 0$
$\implies (a-b)(\sqrt{a}-\sqrt{b})\ge 0$ which is true because $a \gt b$ means both $(a-b)$ and $(\sqrt{a}-\sqrt{b})$ are positive so their product is positive too.
And if $a<b$ then both $(a-b)$ and $(\sqrt{a}-\sqrt{b})$ are negative so their product is positive.
Equality holds if and only if $a=b$.
Method 2:
Here is another method: (It's basically the same as first method but uses the Rearrangement inequality)
Rearrangement inequality states that given two real sequences $a_1\le a_2\le \cdots \le a_n$ and $b_1\le b_2\le \cdots \le b_n$ ,
$a_1b_1+a_2b_2+\cdots+a_nb_n \ge a_1b_{\sigma(1)}+a_2b_{\sigma(2)}+\cdots+a_nb_{\sigma(n)} \ge a_1b_n +a_2b_{n-1}+\cdots a_nb_1$.
In this case $n=2$, WLOG let $a\ge b \implies \sqrt{a}\ge \sqrt{b}$
$\implies a\sqrt{a}+b\sqrt{b} \ge a\sqrt{b}+b\sqrt{a}$
$\implies \frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}} \ge \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}$
$\implies \frac{a}{\sqrt{b}} +\frac{b}{\sqrt{a}} \ge \sqrt{a}+\sqrt{b}$