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The problem is: By choosing a suitable function f prove that

$$\left(t_1 \cdot t_2 \cdot t_3\cdots t_n\right)^\frac{1}{n}\leq\frac{t_1+t_2+t_3+\cdots+t_n}{n}$$

I think that the function is $e^x$ but I don't prove it right.

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    Why don't you show your attempted proof with $e^x$, which BTW is a good choice. – 2017-02-02
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    I edited your question to introduce latex formatting. In future questions try to use the MathJax formatting from the beginning. – 2017-02-02
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    Also that works for $t_i$ non-negative... – 2017-02-02

3 Answers 3

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Note that $\ln x$ is concave and increasing in $x>0$, which follows from calculus. So we have that $$\ln \frac{x_1+x_2+\dots+x_n}{n} \ge \frac{1}{n} \left(\ln x_1+\ln x_2+\dots+\ln x_{n}\right)=\ln (x_{1}x_2x_3 \dots x_n)^\frac{1}{n}$$ As $\ln x$, is increasing we have that $$\frac{x_1+x_2+\dots+x_n}{n} \ge (x_{1}x_2x_3 \dots x_n)^\frac{1}{n}$$We could do a similar thing with $e^x$, by letting $e^{a_{1}}=x_{1}, e^{a_{2}}=x_{2}, \dots, e^{a_{n}}=x_{n}$.

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This is just AM-GM inequality with n variables. For a proof with Jensen's see here.

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Consider the function $f(x) = \ln x$, $f''(x) = -x^{-2} < 0$. Thus $f$ is concave and you get the proof.