Inequality of functions involving logarithm

Question

Define $f(x) = (1+x)\log(1+x)-x$ for $x>-1$. I want to prove that $$ f(x)\geq \dfrac{x^2}{2+2x/3},\quad x\geq 0. $$ I got that this inequality is true if and only if $$ 1\geq g(x):= \exp\left(\dfrac{\frac{5}{3}x^2+2x}{\frac{2}{3}x^2+\frac{8}{3}x+2}\right)-x,\quad x\geq 0. $$ By ploting $g$ I know that $1\geq g (x)$ for all $x\geq 0$, but I don't know how to prove it. Derivating $g(x)$ and try to see if it is decreasing doesn't seems like a promising procedure.

Any idea on how to do it?

Answer 1

I think it's better to make the following:

We need to prove that: $$(1+x)\ln(1+x)\geq\frac{3x^2}{2x+6}+x$$ or $$\ln(1+x)\geq\frac{5x^2+6x}{2(x+3)(x+1)}.$$ Now, let $g(x)=\ln(1+x)-\frac{5x^2+6x}{2(x+3)(x+1)}.$

Thus, $$g'(x)=\frac{x^3}{(x+1)^2(x+3)^2}\geq0,$$ which says $$g(x)\geq g(0)=0$$ and we are done!

Now we see that $$f(x)\geq\frac{x^2}{2+\frac{2}{3}x}$$ for any $x>-1$.

Answer 2

We have $$f(0)\geq \frac{0^2}{2+2\cdot0/3}$$ and the inequality holds if we have $$f'(x)=\log(1+x)\geq\frac{3x(6+x)}{2(3+x)^2}=\frac{d}{dx}\left(\frac{x^2}{2+2x/3}\right)$$ for $x\geq 0$. This is true for $x=0$ and then differentiating once more we get: $$\frac{1}{1+x}\geq\left(\frac{3}{3+x}\right)^3$$ which should hold true for $x\geq0$.