2
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ Hint: using the first result for $x_i = \sqrt{a_i/(a_1+...+a_n)}$. $\endgroup$
    – Khue
    Mar 26 '16 at 8:01
  • $\begingroup$ 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? $\endgroup$
    – Mary Star
    Jan 14 '21 at 9:35
  • 1
    $\begingroup$ @MaryStar Yes that's what I meant. $\endgroup$
    – Khue
    Jan 14 '21 at 9:38
  • $\begingroup$ Ok! Thank you!! :-) @Khue $\endgroup$
    – Mary Star
    Jan 14 '21 at 9:46
  • $\begingroup$ @MaryStar You are welcome! $\endgroup$
    – Khue
    Jan 14 '21 at 9:47
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.