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!
 A: Since the inequality is homogeneous and invariant upon swapping the variables, we may assume that $b=1$ and $a \ge 1$. Then it remains to show that
$$f(a) = \frac{1}{4}(a-1)\log(a) - (\sqrt{a} - 1)^2 \ge 0.$$
Notice that $f(1) = 0$. Therefore we are done if we can show that $f$ is increasing.
Differentiating gives
$$f'(a) = \frac{1}{4} \log(a) + \frac{1}{4} \frac{a-1}{a} - \frac{\sqrt{a} - 1}{\sqrt{a}}$$
and $f'(1) = 0$. Thus we are done if we can show that $f'$ is increasing. Differentiating once more gives
$$f''(a) = \frac{1}{4a} + \frac{1}{4a^2} - \frac{1}{2a\sqrt{a}}.$$
Now $f''(a) \ge 0 \Leftrightarrow a + 1 - 2 \sqrt{a} \ge 0 \Leftrightarrow (a+1)^2 \ge 4a \Leftrightarrow (a-1)^2 \ge 0$, which is true.
A: Wlog a>b, then by Cauchy-Swartz $$\left(\int_b^a \  1\cdot\frac{1}{\sqrt{x}}\  dx\right)^2\leq\int_b^a \  1\ dx \cdot\int^a_b \frac{1}{x}\ dx$$
A: It's not restictive to assume $a>b$, so we can write $a=e^{2s}$, $b=e^{2t}$, with $s>t$ and the inequality to prove becomes
$$
4(e^s-e^t)^2\le(e^s-e^t)(e^s+e^t)(2s-2t)
$$
or
$$
\frac{e^s-e^t}{s-t}\le\frac{e^s+e^t}{2}
$$
By Lagrange's theorem, we know that
$$
\frac{e^s-e^t}{s-t}=e^u
$$
for some $u$, $t<u<s$, so we want
$$
e^u\le \frac{e^s+e^t}{2}
$$
which is true for all $u\in(t,s)$ by the convexity of the exponential function.
A: 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 
$$\frac{e^{2z}-1}{e^{2z}+1}=\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}{2(1+\frac{z^2}{2!}+\frac{z^4}{4!}+...)}\leq{z}.$$
A: Inequality that is another form of a well-known inequalities:
For $0<y<x$
$$ \sqrt{xy}<\frac{x-y}{\ln x-\ln y} < \frac{x+y}{2}. (1)$$
For $a>b$ inequality of enunciation transform so:
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(a-b)(\ln a-\ln b)<=>$$
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})(\ln a-\ln b)<=>$$
$$\sqrt{a} - \sqrt{b} < \frac{1}{2}(\sqrt{a}+\sqrt{b})(\ln \sqrt{a}-\ln \sqrt{b})<=>$$
$$\frac{\sqrt{a} - \sqrt{b} }{\ln \sqrt{a}-\ln \sqrt{b}}< \frac{\sqrt{a}+\sqrt{b}}{2}$$
which is obtained from (1) for $x=\sqrt{a}, y=\sqrt{b}.$
