A proof for AM-GM inequality (Apostol: Mathematical Analysis)

Find the maxima of $(x_1x_2\ldots x_n)^2$ under the constraint $x_1^2+x_2^2+\ldots+x_n^2=1$. Using this result prove that for positive numbers $a_1,a_2,\ldots,a_n$ $$(a_1a_2\ldots a_n)^{1/n} \leq \frac{a_1+a_2+\ldots +a_n}{n}$$

My Answer: I solved the first part using Lagrange multipliers and the maximum is attained at $x_1=x_2=\ldots =x_n =\frac{1}{\sqrt{n}}$.

The part I need help with is how do I use this information to prove the AM-GM inequality?

• Hint: using the first result for $x_i = \sqrt{a_i/(a_1+...+a_n)}$.
– Khue
Mar 26 '16 at 8:01
• Hello @Khue ! Let $\displaystyle{x_k = \sqrt{\frac{a_k}{a_1+\ldots +a_n}}= \sqrt{\frac{a_k}{\displaystyle{\sum_{k=1}^na_k}}}}$. It holds that $\displaystyle{\sum_{k=1}^nx_k^2=\sum_{k=1}^n\frac{a_k}{\displaystyle{\sum_{k=1}^na_k}}=\frac{1}{\displaystyle{\sum_{k=1}^na_k}}\cdot \sum_{k=1}^na_k=1}$ since this condition is satisfied we can use the first result? Jan 14 '21 at 9:35
• @MaryStar Yes that's what I meant.
– Khue
Jan 14 '21 at 9:38
• Ok! Thank you!! :-) @Khue Jan 14 '21 at 9:46
• @MaryStar You are welcome!
– Khue
Jan 14 '21 at 9:47

Hint plug values of all a's as $1/\sqrt{n}$ so you get product as $\frac{1}{\sqrt{n}}$ now we get summation also as $\frac{1}{\sqrt{n}}$ so at maxima of product sum achieves its minimum thus AM-GM inequality is proved