# Applying Jensen's inequality to show $\sqrt[n]{y_1…y_n} \ \leq \frac{y_1+…+y_n}{n}$

(1) Prove $$\sqrt[n]{y_1...y_n} \ \leq \frac{y_1+...+y_n}{n}$$ where $y_1, ..., y_n\in (0,\infty)$

My attempt: I was thinking of using Jensen's inequality with the convex $\phi(t)=e^t, \ X={x_1, ..., x_n},\ \mu(x_i)=\frac{1}{i}$. Am I on the right track? I'm also not sure how to proceed from this.

(2) Prove $$y_1^{\alpha_1}...y_n^{\alpha_n} \ \le \alpha_1 y_1 + ...+ \alpha_ny_n$$ where $\sum_{i=1}^{n}\alpha_i = 1$

## 2 Answers

Take $\log$ both sides and use concavity for $\log$ that is

$$\frac {\sum \log x_i}{n}\le \log \left( {\frac{\sum x_i}{n}}\right)$$

• Thanks! Would you mind to elaborate a little bit more? – wtnmath Apr 4 '18 at 11:20
• @wtnmath by Jensen inequality for a concave function $$\frac {\sum f( x_i)}{n}\le f \left( {\frac{\sum x_i}{n}}\right)$$ and \log x is indeed concave since $$(\log x)''=-\frac1{x^2}<0$$ – gimusi Apr 4 '18 at 15:40

Since $$y=\ln(x)$$ is strictly concave we have $$\ln\sum_{i=1}^{n}\alpha_ia_i\geq \sum_{i=1}^{n}\alpha_i\ln(a_i)=\ln\prod_{i=1}^na_i^{\alpha_i}$$ and since $y=\ln(x)$ is strictly increasing then follows $$\sum_{i=1}^{n}\alpha_ia_i\geq \prod_{i=1}^{n}a_i^{\alpha_i}$$ For the proove of the inweighted $AM-GM$ substitute $$\alpha_i=\frac{1}{n}$$

• Is there a way that I can make use of inequality of (1) to prove (2)? – wtnmath Apr 4 '18 at 15:13
• what do you mean with (1),(2) exactly? – Dr. Sonnhard Graubner Apr 4 '18 at 15:40
• As in using $\sqrt[n]{y_1...y_n} \ \leq \frac{y_1+...+y_n}{n}$ to show $$y_1^{\alpha_1}...y_n^{\alpha_n} \ \le \alpha_1 y_1 + ...+ \alpha_ny_n$$ – wtnmath Apr 4 '18 at 15:54
• I think that is not possible, since your last inequality is the general case – Dr. Sonnhard Graubner Apr 4 '18 at 15:56