# The inequality $( \ln x - \ln y )(x - y) \geq 4(\sqrt{x} - \sqrt{y})^2$

I read somewhere that $$( \ln x - \ln y )(x - y) \geq 4(\sqrt{x} - \sqrt{y})^2$$ for positive $$x, y$$ and would like to prove it. The problem narrows down to showing that the function $$f : (0,1) \to \mathbb{R}$$ defined by

$$f(v) := \int_0^1 \frac{1}{t(1-v)^2 + (1-t) v^2} dt$$

obtains its minimum value at $$v = \frac{1}{2}$$. It would suffice to show $$f$$ is convex since $$f'(1/2) = 0$$ by the symmetry $$f(1-v) = f(v)$$, but convexity of $$f$$ does not seem obvious.

Since our inequality is symmetric, we can assume that $$x\geq y$$, $$x=t^2y,$$ where $$t>1$$ because for $$t=1$$ our inequality is true.
Thus, we need to prove that $$(t^2-1)\ln{t}\geq2(t-1)^2$$ or $$f(t)\geq0,$$ where $$f(t)=\ln{t}-\frac{2(t-1)}{t+1}.$$ Indeed, $$f'(t)=\frac{1}{t}-\frac{4}{(t+1)^2}=\frac{(t-1)^2}{t(t+1)^2}>0,$$ which gives $$f(t)\geq f(1)=0$$ and we are done.
By homogeneity, it suffices to show $$(\log x)(x-1)\geq 4(\sqrt{x}-1)^{2}$$ for $$x>1$$, it suffices to show $$(\log x)(\sqrt{x}+1)\geq 4(\sqrt{x}-1)$$, equivalently, $$(\log\sqrt{x})(\sqrt{x}+1)\geq 2(\sqrt{x}+1)$$, it suffices to show $$(\log x)(x+1)\geq 2(x-1)$$, now let $$\varphi(x)=(\log x)(x+1)-2(x-1)$$, $$\varphi(1)=0$$, but $$\varphi'(x)=1/x+\log x-1$$, $$\varphi'(1)=0$$, but $$\varphi''(x)=-1/x^{2}+1/x=(1/x)(1-1/x)>0$$ for $$x>1$$, so $$\varphi'(x)\geq 0$$ and hence $$\varphi(x)\geq 0$$.
Ignoring the case $$x = y$$, we can assume without loss of generality that $$x > y > 0$$.
Because the function $$f(t) = 1/t$$ is convex: $$\ln\sqrt{x} - \ln\sqrt{y} = \int_{\sqrt{y}}^{\sqrt{x}} f(t)\,dt > (\sqrt{x} - \sqrt{y}) \cdot f\left(\frac{\sqrt{x} + \sqrt{y}}{2}\right),$$ and the result follows upon multiplication of both sides by $$2(x - y)$$.