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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.

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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.

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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 $\varphi(x)\geq 0$.

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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 $2(x - y)$.

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