Bounding $t \ln(1 +t)$ for $t > 0$

Question

How would you derive the following bound? $$t\ln(1+t) \geq 2t - 2\ln(1+t)\qquad \text{for $t>0$.}$$ I have only been able to derive $$t\ln(1+t) \geq t - \ln(1+t)\qquad \text{for $t>0$.}$$

Answer 1

consider the function $$f(t)=\log(1+t)-\frac{2t}{t+2}$$ and $$t>0$$ and we get $f(0)=0$ so we have $$f'(t)=\frac{t-1}{(1+t)^2}$$ so $f'(t)>0$ if $t>1$

Answer 2

The inequality $$t\ln(1+t) ≥ 2t - \ln(1+t)$$ does NOT hold for $t\geq 0$. Take $t=e-1>0$, then $$e-1=(e-1)\ln(e)\geq 2(e-1) - \ln(e)=2e-3\implies 2\geq e$$ which is false! On the other hand, by letting $$f(t)=t\ln(1+t) -2t + 2\ln(1+t)$$ we have that $$f'(t)=\ln(1+t)+\frac{t}{1+t}-2+\frac{2}{1+t}=\frac{(1+t)\ln(1+t)-t}{1+t}$$ which is non-negative by your second bound (the one that you proved!). Hence $f$ is increasing in $[0,+\infty)$ and $f(t)\geq f(0)=0$ implies that for $t\geq 0$, $$t\ln(1+t) ≥ 2t - 2\ln(1+t).$$