To prove that all the terms in the LHS are equal when A.M=G.M...

Question

I have seen in numerous examples that in the inequality of $\dfrac{x_1+x_2+x_3+\cdots+x_{n}}{n}\geq\sqrt[n]{x_1x_2x_3\cdots x_n}$ when the equality holds, $x_1=x_2=x_3=\cdots=x_{n}$ would be true. I knew it for sure when it came to two terms because I could prove that using quadratic equations such as: $$x_1+x_2= 2\sqrt{x_1x_2}\Rightarrow x_1^2+x_2^2+2x_1x_2-4x_1x_2=(x_1-x_2)^2=0\Rightarrow x_1=x_2$$. I also thought that I could prove using binomials for number of terms more than 2, but I couldn't manage to find a general way to prove it for all $n\geq 2$ where $n\in\mathbb{N}$. Perhaps induction might work, but I can't get away with the algebra, what are your suggestions?

Answer 1

Note that:

$$\frac {\sum x_i}{n}\geq\sqrt[n]{\prod x_i}\iff\log \left({\frac {\sum x_i}{n}} \right)\geq \frac{\sum \left(\log x_i\right)}{n}$$

and for convexity properties equality holds if and only if: $$x_1=x_2=...=x_n$$

Answer 2

Another way to prove it is by Lagrange's multipliers.

Consider the function:

$$F(x_i)=\frac {\sum x_i}{n}-\sqrt[n]{\prod x_i}\geq0$$

and suppose wlog

$${\prod x_i}=1\implies F(x_i)=\frac {\sum x_i}{n}-1\geq0$$

the condition for the minimum of F is given by

$$\frac{\partial F}{\partial x_i}=\frac1n=\frac{\lambda}{x_i}\implies x_1=x_2=...=x_n=1$$