How can we show that $e^{-2\lambda t}\lambda^2\le\frac1{e^2t^2}$ for all $\lambda,t\ge0$?

Question

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?

Answer 1

The line $y=x$ is tangent on $y=e^{x-1}$ in $x=1$ and since $e^{x-1}$ is convex, then it lies upper that any tangent of it, specially $x\le e^{x-1}$.

Answer 2

If i take the ln on both sides we get $$-2\lambda t +2\ln(\lambda)\le -2-2\ln(t)$$ and this is not the result given above!