(Solved): In this exercise you will prove the Cauchy-Schwarz inequality for a general inner product space (V ...

In this exercise you will prove the Cauchy-Schwarz inequality for a general inner product space $$(V,??,??)$$. The Cauchy-Schwarz inequality says that $$??\vec{x},\vec{y}????\vec{x}???\vec{y}?$$ for all $$\vec{x},\vec{y}?V$$, with equality if and only if $$\vec{x}$$ and $$\vec{y}$$ are linearly dependent. (a) Show that $$??\vec{x},\vec{y}??=?\vec{x}???\vec{y}?$$ if $$\vec{x}$$ and $$\vec{y}$$ are linearly dependent. (b) Fix $$\vec{x},\vec{y}?V$$, and assume that $$\vec{x}$$ and $$\vec{y}$$ are linearly independent. Show that the function $$P(t)=?\vec{x}?t?\vec{y}{?}^2$$ (a function of $$t$$ ) is a quadratic polynomial in $$t$$ and that the leading coefficient it nonzero. (c) Show that $$P(t)$$ has no real roots. (d) Use the reminder about discriminants below to compute the discriminant of $$P(t)$$ and complete the proof of the Cauchy-Schwarz inequality. Discriminants of quadratic polynomials: the discriminant of a quadratic polynomial with real coefficients $$f(t)=a{t}^2+bt+c?\mathbb{R}[t]$$ is defined to be $${?}_f=b^2?4ac$$. Assuming that $$a?=0$$, the discriminant is related to the roots of $$f$$ as follows. $$\begin{array}{l}{?}_f>0?f(t)\text{ has two distinct real roots, }\\ {?}_f=0?f(t)\text{ has one real root (then it is a double root), and }\\ {?}_f<0?f(t)\text{ has no real roots. }\end{array}$$