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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?

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I think you actually are looking for the AM-GM inequality (https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means) instead. There are easy counter-examples to the inequality you mention. The exponential or the logarithm function is generally used to move between additive and multiplicative means.

Define $a_i=\exp x_i$

(What condition makes sure that we can always write this step?)

Use the Jensen's inequality here:

$$ \exp (\frac 1 n \sum x_i) \leq (\frac 1 n \sum (\exp x_i))$$

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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, $$(a_1a_2\cdots a_n)^{1/n} = \exp{\left(\dfrac{x_1 + \cdots+x_n }{n}\right)} \leq \dfrac{e^x_1 +\cdots+ e^x_n}{n} =\dfrac{a_1 + a_2 + \cdots+a_n }{n}$$

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Since $f(x)=e^x$ is a convex function, by Jensen we obtain: $$\frac{a_1+a_2+...+a_n}{n}=\frac{e^{\ln{a_1}}+e^{\ln{a_2}}+...+e^{\ln{a_1}}}{n}\geq$$ $$\geq e^{\frac{\ln{a_1}+\ln{a_2}+...+\ln{a_n}}{n}}=e^{\ln\sqrt[n]{a_1a_2...a_n}}=\sqrt{a_1a_2...a_n}.$$

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