Help proving the geometric-arithmetic mean inequality using minimization

Question

The problem asks to prove the arithmetic-geometric mean inequality $(a_{1}a_{2}...a_{n})^{1/n} \leq \frac{1}{n}(a_{1}+a_{2}+...+a_{n})$ for $a_{i}>0$ by minimizing the function $f(\vec{x})=\frac{1}{n}(x_{1}+x_{2}+...+x_{n})$ on the surface $g(\vec{x})=x_{1}x_{2}...x_{n}-1=0$.

I found that the minimum occurs when all $x_{i}=1$ so $f(1,...,1)=1$. So now we have $\frac{1}{n}(x_{1}+...+x_{n}) \geq1$ if $x_{1}x_{2}...x_{n}=1$.

I think my work up to this point is correct., but I'm really not sure how to use this fact to prove the G-A inequality. I thought about putting $x_{i}$ in terms of $a$ somehow so their product is always $1$ but I'm just not getting it.

Answer 1

In general given arbitrary positive numbers, $y_1$, $y_2, \ldots, y_n$.

Define $z_i = \frac{y_i}{\left(\prod_{k=1}^n y_k\right)^\frac1n}$

then we have $\prod_{i=1}^n z_i =1$, by your conclusion,

$$\frac{1}{n}\sum_{i=1}^n z_i \ge 1$$

which is $$\frac{1}{n}\sum_{i=1}^n y_i \ge \left(\prod_{i=1}^ny_i\right)^\frac1n$$

Answer 2

Hint: put $x_i = a_i / G$, where $G = (a_1 \cdots a_n)^{1/n}$. It will satisfy the constraint $g(x)=1$.