I am trying to solve the following problem in Kolmogorov's analysis textbook.

Verify that $$ (\sum\limits_{k=1}^n a_k b_k)^2 = \sum\limits_{k=1}^n a_k^2 \sum\limits_{k=1}^n b_k^2 - \frac{1}{2} \sum\limits_{i=1}^n \sum\limits_{j=1}^n (a_i b_j - b_i a_j)^2. $$ Deduce the Cauchy-Schwarz inequality, $$ (\sum\limits_{k=1}^n a_k b_k)^2 \leq \sum\limits_{k=1}^n a_k^2 \sum\limits_{k=1}^n b_k^2. $$ from this identity.

The initial inequality is giving me some trouble, as there does not appear to be an algebraic trick. I have tried expanding out the summand on the right and breaking apart the sum. Since the left-hand is a summation only over $k$, it must be the case that the $i$ and $j$ sums vanish somehow. There doesn't seem to be a good way to factor the right-hand side in order to make this true.

Any help on this would be greatly appreciated.

  • $\begingroup$ Do you mean the initial "equality"? If all else fails, really just multiply it out. It might be a bit difficult in generality at first but then try concrete numbers, e.g. $n = 2$. You will then see what you should have to do. Generalize. $\endgroup$
    – Qi Zhu
    Sep 9, 2019 at 2:29
  • $\begingroup$ use induction to prove the first equality, and you are done. Alternatively expand both sides and finish $\endgroup$
    – Masacroso
    Sep 9, 2019 at 2:29

1 Answer 1


The right-hand side of the first equality is \begin{align*} (RHS)&=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\sum_i\sum_j(a_ib_j-b_ia_j)^2\\ &=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\sum_i\sum_j(a_i^2b_j^2-2a_ib_ia_jb_j+b_i^2a_j^2)\\ &=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\sum_i\sum_ja_i^2b_j^2+\sum_i\sum_ja_ib_ia_jb_j-\frac{1}{2}\sum_i\sum_jb_i^2a_j^2\\ &=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\left(\sum_ia_i^2\right)\left(\sum_jb_j^2\right)+\left(\sum_i a_ib_i\right)\left(\sum_ja_jb_j\right)-\frac{1}{2}\left(\sum_ib_i^2\right)\left(\sum_ja_j^2\right) \end{align*} Change all summation indices to $k$: \begin{align*} (RHS)&=\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)-\frac{1}{2}\left(\sum_ka_k^2\right)\left(\sum_kb_k^2\right)+\left(\sum_k a_kb_k\right)\left(\sum_ka_kb_k\right)-\frac{1}{2}\left(\sum_kb_k^2\right)\left(\sum_ka_k^2\right)\\ &=\left(\sum_ka_kb_k\right)^2 \end{align*}

  • $\begingroup$ Thank you for the incredibly thorough answer. My work for the first several lines was nearly identical, but I couldn't finish it. If I could ask a quick follow-up: why are we able to change the summation indices to $k$? Is it because $i$ and $j$ are just "dummy variables" and it doesn't matter what we call them? If this is the case, it confuses me that we even needed to introduce $i$ and $j$ in the statement of the problem. $\endgroup$
    – user465188
    Sep 9, 2019 at 4:00
  • 1
    $\begingroup$ @Matt.P Precisely. The usage of variables $i$,$j$,$k$ is similar to the use of variables $x,y,z$ for functions. Think about this: take a function $f(\cdot,\cdot)$ from $\mathbb{R}^2$ to $\mathbb{R}$. We usually say that $f$ is a "function of two variabels". The variables themselves do not matter, so whenever we are working with $f$ we can use expressions like "$f(x,y)$ or $f(a,b)$, which will give the same result: If a property $P(x,y)$ is valid for all $x$ and all $y$, then the property $P(a,b)$ is valid for all $a$ and all $b$ (cont.) $\endgroup$ Sep 9, 2019 at 16:43
  • 1
    $\begingroup$ However, we shouldn't write it as "$f(x,x)$", since this implies that the arguments of $f$ are the equal. Another similar case, is when we integrate functions of 2 variables: We can write $\int \int f(x,y)dxdy$, but not $\int \int f(x,x)dxdx$. But we can change the variables if necessary: $\int g(x)dx=\int g(s)ds$. The same is true for summations: You can change indices for summations at will, as long as the same index does not appear in two iterated summations (one inside the other). This is why $i$ and $j$ are necessary in the statement (although "$k$ and $p$" would work just as well). $\endgroup$ Sep 9, 2019 at 16:46

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