Cauchy's inequality is given by: for real numbers, $a_1,...,a_n$, $b_1,...,b_n$, $(a_1^2,...,a_n^2)^{1/2}(b_1^2,...,b_n^2)^{1/2} \geq |a_1b_1+a_2b_2+...+a_nb_n|$. Assuming this, prove that $\sum_{k=1}^n a_k^2 \geq (\sum_{k=1}^n a_k)^2/n$. I tried expanding this inequality to get some resemblance of the Cauchy inequality, but can't seem to get past a certain step. So we have,

$$ a_1^2+...+a_n^2 \geq (a_1+...+a_n)^2/n \\ (a_1^2+...+a_n^2)(b_1^2+...+b_n^2) \geq (a_1+...+a_n)^2(b_1^2+...+b_n^2)/n \\ (a_1^2+...+a_n^2)^{1/2}(b_1^2+...+b_n^2)^{1/2} \geq (a_1+...+a_n)(b_1^2+...+b_n^2)^{1/2}/n^{1/2} $$

where $(b_1,...,b_n)$ is a sequence of reals. We have the left side of the Cauchy inequality and I want to get the right side to resemble the right side of the Cauchy inequality which I do not see as possible with the $n$ term, but we can show that it is less than $|a_1b_1+...+a_nb_n|$ to prove our inequality. However, I am having difficulty doing this as I don't know how to manipulate the right side of the Cauchy inequality to get something useful. Any help would be greatly appreciated. Thanks!


1 Answer 1


How about using $b_k = 1$ for $k = 1,2,\cdots ,n$ in $$(a_1^2+...+a_n^2)(b_1^2+...+b_n^2) \geq (a_1b_1+...+a_nb_n)^2$$

So, $$n(a_1^2+...+a_n^2) = (a_1^2+...+a_n^2)\underbrace{(1^2+...+1^2)}_{n \textrm{ terms }} \geq (a_1+...+a_n)^2$$

as required.

  • $\begingroup$ So this would be a manipulation on the Cauchy inequality? $\endgroup$
    – RXY15
    Jan 23, 2015 at 3:26
  • $\begingroup$ @RXY15 we are using a particular case where $b_k=1$ for each $k=1,2,\cdots,n$ in Cauchy Schwarz Inequality, not manipulate it further. $\endgroup$
    – sciona
    Jan 23, 2015 at 3:29
  • 1
    $\begingroup$ Ahh, OK I see, and then dividing through by $n$ will give the desired result. Thank you for showing me a different approach! $\endgroup$
    – RXY15
    Jan 23, 2015 at 3:38

You must log in to answer this question.

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