How can we show that $$e^{-2\lambda t}\lambda^2\le\frac1{e^2t^2}\tag1$$ for all $\lambda,t\ge0$?
Applying $\ln$ to both sides yields that $(1)$ should be equivalent to $$t\lambda\le e^{t\lambda-1}\tag2.$$ So, if I did no mistake, it should suffice to show $x\le e^{x-1}$ for all $x\ge0$. How can we do this?