# Can the Cauchy Schwarz be reversed, having $\|a\|^2\|b\|^2\le K \langle a,b\rangle^2$?

Suppose $$a_1,a_2,...,a_n,b_1,b_2,...,b_n$$ are real numbers. Does there exist a constant $$K$$ (depending on $$n$$ maybe, but not on $$a_i$$ or $$b_i$$) such that $$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\leq K(a_1b_1+a_2b_2+...+a_nb_n)^2$$?

I don't know how to proceed about it. Any approach/idea is appreciated.

I am basically trying to see whether the well-known Cauchy Schwarz inequality can be reversed in some way. Actually, I am interested in the following problem: does a bound on $$|u'v|$$ imply any bound on $$||u||.||v||$$? Here $$u,v$$ are vectors. This is equivalent to the one I mentioned.

It is impossible: if the vectors $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ are orthogonal (for the standard inner product), that would imply the vectors are $0$.

No, because for example if $a_1=1=b_2$, $b_1=0=a_2$, you would have $$(1+0)(0+1) \leqslant K(0+0)^2,$$ which is false.

The answer is NO. For example, $$(a_1,a_2)=(1,0)$$ and $$(b_1,b_2)=(0,1)$$.

However, if the vectors $$(a_1,\ldots,a_n),(b_1,\ldots,b_n)$$, instead of being arbitrary, LIVE is a cone of angle $$0\le\vartheta<\pi/4$$, then $$(a_1,\ldots,a_n)\cdot(b_1,\ldots,b_n)\ge \cos(2\vartheta\,)(a_1^2+\cdots a_n^2)^{1/2}(b_1^2+\cdots b_n^2)^{1/2}$$

It can be reversed in the following way:

Assume $$0 < a_\min \le a_i \le a_\max$$ and $$0< b_\min \le b_i \le b_\max$$, then $$\|a\|\cdot\|b\|\le K\langle a,b\rangle$$, where $$K = \tfrac{1}{2}(c + c^{\texttt{-}1})$$ and $$c = \sqrt{\frac{a_\max \cdot b_\max}{a_\min\cdot b_\min}}$$

This is known as Pólya-Szegö’s inequality.

• Do you know a necessary and sufficient condition for it to be an equality ? Jul 19 '21 at 11:51
• @P.Quinton if $a$ and $b$ are both the same constant vector then we have equality; if this is also a necessary condition for equality I don't know for sure, but I would suspect so. Jul 19 '21 at 12:22