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The way I learned it, the scalar product is defined the following way:

If $V$ is a vector space, the scalar product is a function $B:V \times V \to \mathbb{R}$ satisfying for all $u,v,w \in V$ and for all $\lambda \in \mathbb{R}$

1) $B(u,v)=B(v,u)$

2) $B(u+v,w)=B(u,w)+B(v,w)$

3) $B(\lambda u,v)=\lambda B(u,v)$

4) $B(u,u) \geq 0$ with equality if and only if $u$ is the neutral element of $V$ under addition.

Then, if we define the norm a vector by $||v||:=\sqrt{B(v,v)}$, we obtain the Cauchy-Schwarz inequality : $$||u|| \cdot ||v|| \geq |B(u,v)|$$ The quite standard scalar product $B(u,v)= u_1v_1+\cdots + u_nv_n$ (where $u=(u_1,\dots,u_n)$ and $v=(v_1,\dots,v_n)$) gives us the non-trivial inequation $$({u_1}^2+\cdots+{u_n}^2)({v_1}^2+\cdots+{v_n}^2) \geq (u_1v_1+ \cdots + u_nv_n)^2$$ Here is my question : for the above inequality, we defined $B(u,v)= u_1v_1+\cdots + u_nv_n$. But do there exist other ways of defining the scalar product that would give us other interesting inequalities ?

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Here is an example from the classic "Cauchy-Schwarz Master Class" by Michael Steele. (at the risk of posing a new problem as the answer!)

Show for all real numbers $x, y, a, b$, the following is immediate by CS inequality, by defining a suitable inner product: $$(5ax + ay + bx + 3by)^2 \leqslant (5a^2+2ab + 3b^2)(5x^2+2xy+3y^2)$$

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A standard kind of example is to consider the linear space of random variables $X$ such that $\mathbf{E}(|X|^2) < \infty$, modulo a.e. zero functions. This has an inner product $$ \langle X,Y\rangle = \mathbf{E}(XY) $$ Cauchy-Schwarz in this case implies that $$ \operatorname{Cov}(X,Y) \le \operatorname{Var}(X)\operatorname{Var}(Y) $$

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  • $\begingroup$ Note that this is not exactly an example of CS as presented in the question, since in this case $\langle X,X\rangle=0$ doesn't imply that $X$ is zero but only "almost everywhere zero". But +1 anyway, this is really a nice application. $\endgroup$ – Arnaud D. Feb 19 '18 at 14:15
  • $\begingroup$ @ArnaudD. Just quotient the ae zero functions, as in the usual construction of $L^2$ $\endgroup$ – Louis Feb 19 '18 at 15:17
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    $\begingroup$ I know this can be done, but I wasn't sure whether you were implicitly doing this or not. The inequality holds in any case, it only changes the conditions for equality a little bit. $\endgroup$ – Arnaud D. Feb 19 '18 at 15:21
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One stock-standard example is the triangle inequality: $$\|x + y\| \le \|x\| + \|y\|.$$ We can prove this using the identity $x \bullet x = \|x\|^2$, and the linearity of the inner product. \begin{align*} \|x + y\|^2 &= (x + y) \bullet (x + y) \\ &= x \bullet x + x \bullet y + y \bullet x + y \bullet y \\ &= \|x\|^2 + x \bullet y + y \bullet x + \|y\|^2 \\ &\le \|x\|^2 + 2|x \bullet y| + \|y\|^2 \\ &\le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2 \\ &= (\|x\| + \|y\|)^2. \end{align*} Taking the square root of both sides yields the result.

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    $\begingroup$ I don't think this answers the question. To me, the question is about examples of applications of the Cauchy-Schwarz inequality with different scalar products, not general consequences of the Cauchy-Schwarz inequality. $\endgroup$ – Arnaud D. Feb 19 '18 at 11:42

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