0
$\begingroup$

If $a_1,a_2,\dots,a_n \in R^+$, then prove that $$(a_1+a_2+\dots+a_n)^2 \le n^2 (a_1^2+a_2^2+\dots+a_n^2)$$ I attempted to solve this problem but was unable to.I used the Cauchy Schwarz Inequality for the two sets $\{1,1,1, \dots\}$ and $\{a_1,a_2,\dots,a_n\}$ but the result was $$(a_1+a_2+\dots+a_n)^2 \le n(a_1^2+a_2^2+\dots+a_n^2).$$ Could you please help me in solving this problem?

$\endgroup$
3
  • 4
    $\begingroup$ Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :) $\endgroup$
    – mrtaurho
    Commented Oct 4, 2018 at 17:32
  • $\begingroup$ Hint: $n\leq n^2$. $\endgroup$
    – Wojowu
    Commented Oct 4, 2018 at 17:41
  • $\begingroup$ Thank you for the hint $\endgroup$ Commented Oct 4, 2018 at 17:43

1 Answer 1

3
$\begingroup$

While your attempt leads to an answer, as explained in a comment, by noting $n \le n^2$, there is a way that seems more elementary to me:

Let $a_m$ be the largest of the $a_i$, then $$(a_1+a_2+\dots+a_n)^2 \le (na_m)^2 = n^2 a_m^2 \le n^2 (a_1^2+a_2^2+\dots+a_n^2) $$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .