Inequality through calculus To prove $x\cdot \log(x)\geq (x-1)\cdot \log(x+1)$ when  $x\geq 1$
I tried to do it the following way
$f(x)=x\cdot \log(x)-(x-1)\cdot \log(x+1)$
$f''(x)=\frac{1-x}{x(x+1)^{2}}\leq0;$ $x\geq1$
$\therefore f'(1)=-1+\log2> 0
$
So at $x=1$ the function increase but i have to prove the function is always greater than $0$. How do i do that?
 A: Easy way to prove the inequality; $x\geq 1$ therefore $\frac{1}{x}\leq 1$. We use Jensen inequality and we have:
$$
 \frac{1}{x} \log 1+(1- \frac{1}{x})\log(x+1)\leq \log\left( \frac{1}{x}.1+(1- \frac{1}{x})(x+1)\right)=\log x
$$
which gives simply the proof.
A: $f'(x)=\ln\left(\frac{x}{x+1}\right)+\frac{2}{x+1}$; thus
$\lim_{x\to\infty}f'(x)=0$.  Now we know that $f''(x)=0\iff x=1$.  Would $f'$ had a zero there would be another zero of $f''$.  Furthermore $f'(1)>0$.
A: An alternative solution
For $x=1$ the equality is given. So assume $x> 1$ then all terms are positive. So we have
$$ x \cdot \log x \geq (x-1) \cdot \log (x+1) \iff \frac{x}{x-1} \geq \frac{\log(x+1)}{\log(x)}$$
Furthermore we know via mean value theorem
$$\log(x+1)-\log(x) = \log'(\xi)$$ 
where $\xi$ is in $(x,x+1)$. Hence we can rewrite the above inequality into 
$$1+\frac{1}{x-1} \geq 1+ \frac{\frac{1}{\xi}}{\log(x)}$$
Furthermore as $\log(x+1)\leq x$ for $x\geq 0$
$$1+\frac{\frac{1}{\xi}}{\log (x)}\leq 1+\frac{\frac{1}{\xi}}{x-1}$$
So actually our inequality reduces to 
$$1\geq \frac{1}{\xi}$$ where $\xi \in (x,x+1)$, so surely $\xi\geq 1$, which is trivally right.
