<s>[INST] 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!
 [/INST] The equality holds for $a=b$, so we can just consider the case for $a>b$. In this case, to show this inequality, it is equivalent to show that 
$$4\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\leq(\ln(a)-\ln(b)).$$
 As $a>b>0$, we can suppose that $a=e^x$ and $b=e^y$ where $x>y,\ x,y\in\mathbf{R}$. So we have to show that
$$4\frac{e^\frac{x}{2}-e^\frac{y}{2}}{e^\frac{x}{2}+e^\frac{y}{2}}\leq(x-y),$$
or 
$$4\frac{e^\frac{x-y}{2}-1}{e^\frac{x-y}{2}+1}\leq(x-y).$$
Now, denote $\frac{x-y}{4}=z$, the problem becomes to show that for $z>0$ the following inequality holds
$$\frac{e^{2z}-1}{e^{2z}+1}\leq{z}.$$
Obviously, this holds for $z\geq{1}$. So we only consider the case for $0<z<1$, and we have that 
$$\frac{e^{2z}-1}{e^{2z}+1}=\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}{2(1+\frac{z^2}{2!}+\frac{z^4}{4!}+...)}\leq{z}.$$
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