Answer to Question #103712 in Algebra for Sourav

Question

prove that ((x+y+z)/3)^(x+y+x) <=x^x.y^y.z^z<=((x²+y²+z²)/(x+y+z))^(x+y+z) where x,y,z bleongs to N

Answer 1

1) We prove the first of the inequalities

$$\left(\frac{x+y+z}{3}\right)^{x+y+z}\le x^x\cdot y^y\cdot z^z\Longleftrightarrow\\[0.5cm] \ln\left(\left(\frac{x+y+z}{3}\right)^{x+y+z}\right)\le\ln\left(x^x\cdot y^y\cdot z^z\right)\Longleftrightarrow\\[0.5cm] \left.(x+y+z)\ln\left(\frac{x+y+z}{3}\right)\le x\ln x+y\ln y+z\ln z\right|\cdot\frac{1}{3}\\[0.5cm] \boxed{\frac{x+y+z}{3}\ln\left(\frac{x+y+z}{3}\right)\le\frac{x}{3}\ln x+\frac{y}{3}\ln y+\frac{z}{3}\ln z}$$

Now recall Jensen's inequality. For a real convex function $\varphi$ , numbers $x_{1},x_{2},\ldots ,x_{n}$  in its domain, and positive weights $a_{i}$ , Jensen's inequality can be stated as:

$$\varphi\left(\frac{\sum a_ix_i}{\sum a_i}\right)\le\frac{\sum a_i\varphi(x_i)}{\sum a_i}$$

( More information: https://en.wikipedia.org/wiki/Jensen%27s_inequality )

In our case,

$$a_1=a_2=a_3=1\\ \varphi(x)=x\cdot\ln x$$

It remains to prove that the function $\varphi(x)=x\cdot\ln x$ is convex.

To do this, we calculate the second derivative

$$y=x\ln x\rightarrow y'=\ln x+1\rightarrow y''=\frac{1}{x}>0, \forall x\in\mathbb N$$

Q.E.D.