1. **Problem statement:** (i) Show that for any real number $x$ and vectors $u$ and $v$, the equality $$ (u + xv) \cdot (u + xv) = |u|^2 + 2x(u \cdot v) + x^2 |v|^2 $$ holds. (ii) If $A, B, C$ are real numbers such that $$ A + Bx + Cx^2 \geq 0 $$ for any real $x$, explain why $$ B^2 - 4AC \leq 0. $$ (iii) Using (i) and (ii), deduce the Cauchy-Schwarz inequality $$ |u \cdot v| \leq |u| |v|. $$ --- 2. **Step (i): Expand and simplify the dot product** Recall the distributive property of the dot product: $$ (a + b) \cdot c = a \cdot c + b \cdot c. $$ Calculate: $$ (u + xv) \cdot (u + xv) = u \cdot u + u \cdot (xv) + (xv) \cdot u + (xv) \cdot (xv). $$ Since dot product is commutative and scalar multiplication factors out: $$ u \cdot (xv) = x(u \cdot v), \quad (xv) \cdot u = x(v \cdot u) = x(u \cdot v), $$ $$ (xv) \cdot (xv) = x^2 (v \cdot v) = x^2 |v|^2. $$ So, $$ (u + xv) \cdot (u + xv) = |u|^2 + x(u \cdot v) + x(u \cdot v) + x^2 |v|^2 = |u|^2 + 2x(u \cdot v) + x^2 |v|^2. $$ --- 3. **Step (ii): Quadratic form non-negativity implies discriminant condition** Given $$ A + Bx + Cx^2 \geq 0 \quad \text{for all real } x, $$ this quadratic in $x$ never takes negative values. Recall the quadratic formula roots: $$ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2C}. $$ If $B^2 - 4AC > 0$, the quadratic has two distinct real roots, so it takes negative values between roots. If $B^2 - 4AC = 0$, the quadratic touches the $x$-axis at one root and is non-negative. If $B^2 - 4AC < 0$, the quadratic has no real roots and is always positive or always negative. Since the quadratic is non-negative for all $x$, it must be that $$ B^2 - 4AC \leq 0. $$ --- 4. **Step (iii): Deduce the Cauchy-Schwarz inequality** From (i), set $$ A = |u|^2, \quad B = 2(u \cdot v), \quad C = |v|^2. $$ Since $$ (u + xv) \cdot (u + xv) = A + Bx + Cx^2 \geq 0 \quad \text{for all } x, $$ by (ii), $$ B^2 - 4AC \leq 0. $$ Substitute $A, B, C$: $$ (2(u \cdot v))^2 - 4 |u|^2 |v|^2 \leq 0, $$ which simplifies to $$ 4(u \cdot v)^2 - 4 |u|^2 |v|^2 \leq 0. $$ Divide both sides by 4: $$ (u \cdot v)^2 - |u|^2 |v|^2 \leq 0. $$ Rewrite: $$ (u \cdot v)^2 \leq |u|^2 |v|^2. $$ Taking square roots (both sides non-negative): $$ |u \cdot v| \leq |u| |v|. $$ This is the Cauchy-Schwarz inequality. --- **Final answer:** $$ (u + xv) \cdot (u + xv) = |u|^2 + 2x(u \cdot v) + x^2 |v|^2, \quad B^2 - 4AC \leq 0, \quad |u \cdot v| \leq |u| |v|. $$