Prove that for all $x, y>0$, $\ln \left(\frac{x}{y}\right) \geqslant 4 \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}$
Is there any role for mean value theorem in the proof?, Can we use the fact $\ln (x)<\sqrt{x},~\forall x>0$
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Have you tried the substitution $u = \frac{x}{y}$? – Charlie Vanaret Aug 05 '20 at 17:50
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1This inequality does not hold as stated. Note that the inequality reverses when $x$ and $y$ are interchanged. – WimC Aug 05 '20 at 18:08
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Two closely related questions: Prove $\frac{x+1}{x-1}\ln x \geq 2$, prove that $ -2 + x + (2+x)e^{-x}>0 \quad \forall x>0$. – Calum Gilhooley Aug 05 '20 at 20:28
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Let us prove $$\ln u \ge 2 \frac{u-1}{u+1}, u \ge 1$$ Let $$f(u)=(u+1)\ln u-2u+2\implies f'(u)=\ln u+(u+1)/u-2=\ln u+1/u-1$$ $$\implies f''(u)=\frac{u-1}{u^2}.$$ For $u \ge 1$, $f'(u)$ is an increasing function so as $$ u \ge 1 \implies f'(u) \ge f'(0)=0.$$ This means $f(u)$ is an increasing function. Hence $$u \ge 1 \implies f(u) \ge f(1)=0$$ Let $x>y>0$ Let $u=\sqrt{x/y}\ge 1,$ so the required inequality follows.
Z Ahmed
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Hint
Define $u={x\over y}$, hence you need to prove that $$\ln u\ge 4{\sqrt u-1\over \sqrt u+1}$$for $u>0$.
Mostafa Ayaz
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2I think $u=\sqrt{\frac xy}$ is even better because it leads to $$\ln(u)\geq2\frac{u-1}{u+1}$$ – Maximilian Janisch Aug 05 '20 at 17:55
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Use the substitution as mentioned in the comments. Let $u=\sqrt{x/y}$. Then we have to prove that $\ln (u)-2*\frac{u-1}{u+1} \geq 0$. Let $p(u)=\ln (u)-2*\frac{u-1}{u+1}$ Then $$p(u+\delta)-p(u)=\ln \left(\frac{u+\delta}{u}\right)+\frac{2\delta}{(u+\delta+1)(u+1)} \geq 0$$. Hence $p(u)$ is an increasing function and hence greater than $0$. Please note that the condition is $u \geq 1$ or $x \geq y$.
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