Inequality real analysis about logarithm

Question

Can you proof that

$$\forall t \ge -1, (1+t)\log(1+t) -t \ge \frac{t^{2}}{2(1+\frac{t}{3})} $$

A elegant proof will be appreciated !

I tried derivation and power series.

Thanks and regards.

Answer 1

Let $f(t)=(1+t)\ln(1+t)-t-\frac{t^2}{2\left(1+\frac{t}{3}\right)}.$

Thus, $$f''(t)=\frac{t^2(t+9)}{(t+1)(t+3)^3}\geq0.$$ Thus $f'$ increases.

But $$f'(t)=\ln(1+t)-\frac{3t(t+6)}{2(t+3)^2}\rightarrow-\infty$$ for $t\rightarrow-1^+$ and $f'\rightarrow+\infty$ for $t\rightarrow+\infty$, which says that $0$ is an unique root of $f'$,

which gives that for $t=0$ our function gets a minimal value.

Id est, $$f(t)\geq f(0)=0$$ and we are done!