I tried to prove the following inequality:

For $\tau\ge 1$ and $A>0$, $$\frac{1}{\tau} \exp\left(A\right)E_1\left(A\right)\le \exp\left({\tau A}\right)E_1\left({\tau A}\right),$$ where $E_1(x)=\int_x^\infty \frac{\exp(-t)}{t} dt$ denotes the exponential integral.

  • $\begingroup$ It would look nicer if you defined $x=\frac1{A}$ $\endgroup$ – iamvegan Oct 18 '16 at 12:28
  • $\begingroup$ I edited my question according to your suggestion. $\endgroup$ – KJ Choi Oct 18 '16 at 13:02

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