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

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?

## marked as duplicate by Martin R, Community♦Mar 20 at 19:32

• The substitution $u=x-1$ transforms $x \le e^{x-1}$ into $u+1 \le e^u$, which is answered in the above-mentioned Q&A. – Martin R Mar 20 at 17:57
• Note that applying the square root to (1) gives (2), not an application of the logarithm. – Martin R Mar 20 at 18:33
• @MartinR Both is legitimate. I've applied the logarithm, rearranged and then the exponential again. – 0xbadf00d Mar 20 at 18:56

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 $$x\le e^{x-1}$$.
• The question you mentioned in the comments is not a duplicate of this one unless if $e^x-1=e^{x-1}$. A duplication does not make an answer wrong but if I knew the duplication (though I could guess a bit) I wouldn't have answered it unless through comments... – Mostafa Ayaz Mar 20 at 17:53
• The simple substitution $u=x-1$ transforms $x \le e^{x-1}$ into $u+1 \le e^u$ .... – Martin R Mar 20 at 17:55
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!
• Simply divide by $2$, use $\ln(\lambda t)=\ln\lambda+\ln t$ and apply the exponential function again. Then you obtain $(2)$. – 0xbadf00d Mar 20 at 19:31