Given: 0 < a < b, Prove: $\sqrt{b} - \sqrt{a} < \frac{1}{3}(\frac{b}{\sqrt{a}}-\frac{a}{\sqrt{b}})$
As usual, we need to start from one side and reach the other by deduction. However, I am not sure I am going to get there. The solution may require Calculus but not to a highly advanced level.
The graph of $y=\frac{1}{2\sqrt{x}}$ is:
This shows us that the mean value of the function between a and b is less than the mean of f(a) and f(b) as the graph is concave.
In other words:
For the range $[a,b]$,
$f_{mean} \ne \frac{f(a)+f(b)}{2}$
Rewrite like this (divide by $ab$): $${1\over \sqrt{b}} - {1\over \sqrt{a}} < \frac{1}{3}(\frac{1}{a\sqrt{a}}-\frac{1}{b\sqrt{b}})$$ and then $${1\over \sqrt{b}} +\frac{1}{3}\frac{1}{b\sqrt{b}} < \frac{1}{3}\frac{1}{a\sqrt{a}}+{1\over \sqrt{a}}$$ If we put $a'={1\over \sqrt{a}}$ and $b'={1\over \sqrt{b}}$ then we get $0<b'<a'$ and $$3b' + b'^3 <3a'+a'^3$$
Now observe
$$f(x) = 3x+x^3 $$
Since $f'(x) =3+3x^2>0$ the $f$ is strictly increasing and thus conclusion.