### Archive

Archive for the ‘Inequality’ Category

## Prove that e^pi > pi^e , Inequality

November 25, 2011 1 comment

$e^{\pi} > {\pi}^e$

You can use natural logarithm to prove this equality. You will get

$\pi > e \cdot \ln \pi$

$\pi/e > \ln \pi$

$A.M. = \frac{ a_1 + a_2 + a_3 + \ldots + a_n}{n} \: and \: G.M. = \sqrt[n]{ a_1 \cdot a_2 \cdot a_3 \cdots a_n } . Let's \: proof \: A.M. \ge G.M. \: a_i \: is \: non-negative \: Real \: number.$
First of all we need to proof that $(a_1 + a_2 )/2 \ge \sqrt{ a_1 \cdot a_2 }$ It is easy, you just have to know $( \sqrt{a_1} + \sqrt{a_2})^2 \ge 0$