One lower bound for $(1+x)\log(1+x)-x$

Question

Problem

When studying Chernoff bound, one result is used without proof and reference, which is $$ (1+x)\log(1+x)-x\geq \frac{x^2}{\frac{2}{3}x+2} $$ I am wondering how this is proved.

What I Have Done

I checked the minimum of LHS and maximum of RHS, this indeed holds. But when it comes to prove this, this sort of check is far from enough.

Something I think relatable is doing some Taylor expansion of LHS, but I did not get the result.

Could someone help me, thank you in advance.

Edit

Take the second-order derivative of $f(x)=(1+x)\log(1+x)-x-\frac{x^2}{\frac{2}{3}x+2}$ gives us $f''(x)=\frac{x^2\, \left(x + 9\right)}{\left(x + 1\right)\, {\left(x + 3\right)}^3}$, which shows the correctness of the answer below.

Answer 1

Hint: Study the function $f(x) = LHS-RHS$ and show it is non-negative

Detailed hint: $f$ is infinitely differentiable on $(-1,\infty)$. Derive $f$ (twice): $f''$ is easy to handle, as it is a rational function (no more logarithms); it has a single root at $0$ and is always non-negative. This means $f'$ is non-decreasing; since $f'(0)=0$, we have $f$ decreasing on $(-1,0)$ and increasing on $(0,\infty)$. But $f(0)=0$, and thus $f(x)\geq f(0)=0$ for all $x$.

Answer 2

Also, we can make the following.

We need to prove that $f(x)\geq0,$ where $$f(x)=\ln(1+x)-\frac{5x^2+6x}{2(x^2+4x+3)}.$$ We see that $$f'(x)=\frac{x^3}{(x^2+4x+3)^2},$$ which says that $$f(x)\geq f(0)=0$$ and we are done!