Inequality involving logarithm

Question

I've seen the following inequality

$$ 4(\sqrt{x}-\sqrt{y})^2\leq (x-y)(\log(x)-\log(y)) $$

How can I prove something like that? In general, I'm interested in any lower bound for

$$ (x-y)(\log(x)-\log(y)), $$ so, any reference on that is appreciated!

Thanks!

Answer 1

Assuming $x>y>0$, the given inequality just follows from Cauchy-Schwarz:

$$ \int_{y}^{x}\frac{dt}{t}\int_{y}^{x}1\,dt \geq \left(\int_{y}^{x}\frac{dt}{\sqrt{t}}\right)^2.$$

Answer 2

For $x=y>0$ it's obvious.

Since our inequality is symmetric, we can assume that $x=yt$, where $t>1$

and we need to prove that $f(t)\geq0$, where $$f(t)=\ln{t}-\frac{4(\sqrt{t}-1)}{\sqrt{t}+1}=\ln{t}-\frac{4(\sqrt{t}+1-2)}{\sqrt{t}+1}.$$ But, $$f'(t)=\frac{1}{t}-\frac{4}{\sqrt{t}(\sqrt{t}+1)^2}=\frac{(\sqrt{t}-1)^2}{t(\sqrt{y}+1)^2}\geq0,$$ which says $f(t)\geq f(1)=0$