Given a positive numbers $a$ so that $a\neq 1$. Prove that $\frac{\log a}{a-1}\leqq\frac{1+a^{1\div 3}}{a+a^{1\div 3}}$ .

Question

Given a positive numbers $a$ so that $a\neq 1$. Prove that $$\frac{\log a}{a- 1}\leqq \frac{1+ a^{1\div 3}}{a+ a^{1\div 3}}$$

Note $\log a= \log_{e} a$ .

It would be hard if derivative the polynomial $\frac{\log a}{1- a}+ \frac{1+ a^{1\div 3}}{a+ a^{1\div 3}}$ , even Wolfram|Alpha. I need to the helps

Yes, Mr. Michael Rozenberg

Take $x= a^{3}$

$$\frac{{\rm d}}{{\rm d}x} \left ( \frac{1+ x}{x+ x^{3}}- \frac{\log x^{3}}{x^{3}- 1} \right )= \frac{1}{x+ x^{3}}- \frac{3}{(x^{3}- 1)x}- \frac{(3x^{2}+ 1)(1+ x)}{(x+ x^{3})^{2}}+ \frac{3x^{2}\log x^{3}}{(x^{3}- 1)^{2}}= 0$$ I can't solve it

Answer 1

Let $a>1$ and $a=x^3$.

Thus, we need to prove that $f(x)\geq0,$ where $$f(x)=\frac{(1+x)(x^3-1)}{3(x^3+x)}-\ln{x}.$$ But, $$f'(x)=\frac{(x-1)^4(x^2+x+1)}{3(x^3+x)^2}\geq0,$$ which gives $f(x)\geq0.$

The case $0<a<1$ is the same.