An inequality involving e

Question

I need to prove the inequality $e^{2x-\frac{2x^2}{3}} \geq \frac{1+x}{1-x+\frac{2x^2}{3}}$ for all $x \leq 3$. This is true according to Wolfram Alpha. Any ideas?

Answer 1

Note that you are trying to prove that $e^{q(x)} \geq \frac{p(x)}{p(x)-q(x)}$ for all $x$ smaller than the largest $0$ of $q(x)$...

Answer 2

We need to prove that $f(x)\geq0$, where $$f(x)=\frac{2x(3-x)}{3}-\ln(x+1)+\ln\left(1-x+\frac{2x^2}{3}\right)$$ for $-1<x\leq3$ because for $x\leq-1$ our inequality is obviously true.

Indeed, $f'(x)=\frac{8(2-x)x^3}{3(x+1)(2x^2-3x+3)}$, which gives $x_{min}=0$

and since $f(3)=0$, we are done!