Prove that $-\frac{\sqrt{x}}{1+x}\log{x} \leq \log{2}$ for $0 < x < 1$ Graphically and numerically it is obvious but I'm looking for an analytical reasoning.
Just maximizing the left hand side does not yield an analytical expression for the maximum.
I also tried some known bounds for $\log$, but all of them had "overshoot".
 A: Put $y=\frac{1}{\sqrt{x}}$ to get $(y+\frac{1}{y})\log 2-2\log y \geq 0, y>1$.
Differentiating gives $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$, so we have critical points when $(y-\frac{1}{y})\log 2=2$. Solving the quadratic in $y$ and taking the positive root gives $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. It suffices to check the inequality for this value and the end points. ($y \to 1$ and $y \to +\infty$).
Put $y=1$ to get $2\log 2-2 \log(1)$ which is clearly non-negative. If the inequality holds for $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$, then since the gradient $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$ is clearly non-negative when $y>\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$, the inequality holds as $y \to +\infty$.
Finally we prove the inequality when $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. Rewrite the inequality as $2y\log 2 \geq (y-\frac{1}{y})\log 2+2\log y$ and substitute in the value to get $2(1+\sqrt{(\log 2)^2+1}) \geq 2+2\log(\frac{1+\sqrt{(\log 2)^2+1}}{\log 2})$. Put $z=\sqrt{(\log 2)^2+1}$, so we get $z+\log(\log 2) \geq \log (1+z)$, or equivalently, $e^z\log 2 \geq 1+z$. The function $e^x\log 2-(1+x)$ is an increasing function for $x \geq 1$ (The gradient is $e^x\log 2-1 \geq e\log 2-1>0$), and $z>\sqrt{1+0.69^2}>1.2$, so it suffices to show that $e^{1.2}\log 2 \geq 2.2$. Note that $\log 2>0.69$ and $e^{1.2}>1+1.2+\frac{1.2^2}{2}+\frac{1.2^3}{6}=3.208>3.2$, and $3.2(0.69)=2.208>2.2$, so we are done.
A: when you take the derivative you will get such an expression
$$x^{-\frac{\sqrt{x}}{1+x}} \left(-\frac{1}{\sqrt{x} (1+x)}+\left(\frac{\sqrt{x}}{(1+x)^2}-\frac{1}{2 \sqrt{x} (1+x)}\right) \text{Log}[x]\right)$$
Then solving for 
$$\left(-\frac{1}{\sqrt{x} (1+x)}+\left(\frac{\sqrt{x}}{(1+x)^2}-\frac{1}{2 \sqrt{x} (1+x)}\right) \text{Log}[x]\right)=0$$
will give you some roots. one of them is $0.0907763$ and the other one is $11.0161$. If one checks carefully then it can be seen that the root $0.0907763$ yields the maximum which is $1.94011$.
Okay more analytical:
From the equation above we get:
$\log(x)=\frac{\frac{1}{\sqrt{x} (1+x)}}{\left(\frac{\sqrt{x}}{(1+x)^2}-\frac{1}{2 \sqrt{x} (1+x)}\right)}$
If we insert this into 
$$\frac{-\sqrt{x}}{1+x}\log(x)$$
and do some algebra we get
$$-2y\frac{e^y}{1+e^{2y}}$$
where $-1<y<\infty$ and originally $y=\frac{1+x}{x-1}$
Now it is easy to see that this function has a maxima at $-1$ which is less than $\log 2$
