Problem

When studying Chernoff bound, one result is used without proof and reference, which is
$$
(1+x)log(1+x)-xgeq frac{x^2}{frac{2}{3}x+2}
$$
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 $f''(x)=frac{x^2, left(x + 9right)}{left(x + 1right), {left(x + 3right)}^3}$, which shows the correctness of the answer below.

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 $x$.

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!