Interesting inequalities that can be derived from Cauchy-Schwarz 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 ?
 A: 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 
$$
\operatorname{Cov}(X,Y) \le \operatorname{Var}(X)\operatorname{Var}(Y)
$$
A: 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: 
$$(5ax + ay + bx + 3by)^2 \leqslant (5a^2+2ab + 3b^2)(5x^2+2xy+3y^2)$$
A: 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.
