Let $a,b$ be two $n$ dimensional vectors, we want to show that

$(\sum\limits_k a_k b_k)^2 \leq \sum\limits_k b_k a_k^2 \sum\limits_k b_k$

Recall the Cauchy Schwarz inequality is given as

$|\langle x,y \rangle| \leq \|x\|\|y\|$

then let $x = a,y = b$


$|\langle a,b \rangle| \leq \|a\|\|b\|$

= $|a^Tb| \leq \sqrt{a^Ta} \sqrt{b^Tb}$

= $|a^Tb|^2 \leq a^Ta b^Tb$

$\Rightarrow$ $|\sum\limits_k a_k b_k|^2 \leq \sum\limits_k a_k^2 \sum\limits_k b_k^2$

At this point how can we eliminate the absolute value sign for the first term and massage the right hand side to what we wish to show?

  • 1
    $\begingroup$ you need extra conditions that $a_k>0$ and $b_k>0$, otherwise this is not true. $\endgroup$
    – Hamed
    Oct 11, 2015 at 15:56

1 Answer 1


$|x|^2 = (x)^2$. Also,

$$ \left(\sum_k a_k b_k\right)^2 = \left|⟨ \sqrt b· a, \sqrt{b}\rangle \right|^2\overset{C-S}{\leq} \sum_k b_k a_k^2 \sum_k b_k $$

(Note that we need $b\geq 0$.)


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