Connection between $e^{x}$ convexity and inequality Given that  $e^x$ is strictly convex, I was asked to prove that 
$$\sqrt[n]{a_1a_2a_3\cdots a_n} \le \dfrac{a_1 + a_2 + a_3+\cdots +a_n} n, \qquad a_1,a_2,a_3,\ldots,a_n > 0$$
I have not seen the relationship between $e^x$ convexity and the above inequality. How do I connect these two concepts?

You have  typo's - Your cube root should be an nth root and the right side of your inequality should be $\dfrac{a_1 + a_2 + \cdots+a_n }{n} $
 Now there  is no  problem . Let $x_j = \ln (a_j)$  so that $e^x_j = a_j$  and apply the strict convexity,
$$(a_1a_2\cdots a_n)^{1/n} =  \exp{\left(\dfrac{x_1 + \cdots+x_n }{n}\right)} \leq \dfrac{e^x_1 +\cdots+ e^x_n}{n} =\dfrac{a_1 + a_2 + \cdots+a_n }{n}$$
