Prove inequality $(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$ I am trying to prove the following inequality:
$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$
for all $a>0, b>0$.
Does anyone know how to prove it?
Thanks a lot in advance!

Since the inequality is homogeneous and invariant upon swapping the variables, we may assume that $b=1$ and $a \ge 1$. Then it remains to show that
$$f(a) = \frac{1}{4}(a-1)\log(a) - (\sqrt{a} - 1)^2 \ge 0.$$
Notice that $f(1) = 0$. Therefore we are done if we can show that $f$ is increasing.
Differentiating gives
$$f'(a) = \frac{1}{4} \log(a) + \frac{1}{4} \frac{a-1}{a} - \frac{\sqrt{a} - 1}{\sqrt{a}}$$
and $f'(1) = 0$. Thus we are done if we can show that $f'$ is increasing. Differentiating once more gives
$$f''(a) = \frac{1}{4a} + \frac{1}{4a^2} - \frac{1}{2a\sqrt{a}}.$$
Now $f''(a) \ge 0 \Leftrightarrow a + 1 - 2 \sqrt{a} \ge 0 \Leftrightarrow (a+1)^2 \ge 4a \Leftrightarrow (a-1)^2 \ge 0$, which is true.
