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Theorem (Cauchy-Schwarz Inequality) : If $u$ and $v$ are vectors in an inner product space $V$, then $$\langle u,v\rangle ^2\leqslant \langle u,u\rangle \langle v,v\rangle .$$

Proof : If $u=0$, then $\langle u,v\rangle = \langle u,u\rangle=0$ so that the inequality clearly holds. Assume now that $u\neq 0$. Let $a=\langle u,u\rangle$, $b=2 \langle u,v\rangle$, $c=\langle v,v\rangle$, let $t$ be any real number. By the positivity axiom, the inner product of any vector with itself is always non-negative. Therefore $$0\leqslant\langle (tu+v),(tu+v)\rangle =\langle u,u\rangle t^2+2\langle u,v\rangle t+\langle v,v\rangle =at^2+bt+c.$$

This inequality implies that the quadratic polynomial $at^2+bt+c$ has no real roots or a repeated real root. Therefore its discriminant must satisfy $b^2-4ac\leqslant0$. Expressing $a$,$b$ and $c$ in terms of $u$ and $v$ gives $$4\langle u,v\rangle^2-4\langle u,u\rangle \langle v,v\rangle \leqslant 0$$ or equivalently, $$ \langle u,v\rangle^2\leqslant\langle u,u\rangle\langle v,v\rangle.\blacksquare$$

Doubt : How do we know that $at^2+bt+c$ has no real roots or a repeated real root?

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  • $\begingroup$ I think there's a missing assertion that $0<\langle (tu+v),(tu+v)\rangle$ so long as $tu+v\neq 0$. So the inequality is strict in this case, hence the polynomial has no real roots. This also implies $b^2-4ac<0$. Hence, equality in cauchy schwartz can occur iff $u,v$ are linearly dependent. Another approach is to simply minimize the polynomial in $t$, so that after plugging in the minimum $t$, you'll get back cauchy schwartz $\endgroup$ – Alex R. Jun 9 '17 at 17:09
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Since you have $at^2+bt+c\geq 0$, you either have a double root or no real roots. In fact, if a polynomial of degree 2 has two distinct roots, then it is positive on a certain interval and negative on another one (you should check this). Thus, the polynomial $at^2+bt+c$ must not have two distinct roots because it is positive. Also, a polynomial of degree two has no real roots iff it's discriminant is negative.

You can also check that the given polynomial has a double root if and only if $$\langle u,v\rangle^2=\langle u,u\rangle\langle v,v\rangle.$$

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  • $\begingroup$ How do I know the polynomial discriminant is negative given the fact the polynomial is positive? $\endgroup$ – Pedro Gomes Jun 9 '17 at 17:28
  • $\begingroup$ Look at the formula for the solutions of a second degree polynomial: $x = \frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$. There exist real roots when the square root is real, i.e. when $b^2 - 4ac \geq 0$. $\endgroup$ – md2perpe Jun 9 '17 at 17:34
  • $\begingroup$ @md2perpeI know that. My question is How do you know given the fact $at^2+bt+c\geq 0$ that $b^2 - 4ac \leqslant 0$? $\endgroup$ – Pedro Gomes Jun 9 '17 at 17:36
  • $\begingroup$ If $a>0$ and $at^2+bt+c>0$ for all $t$ then the equation $at^2+bt+c=0$ has no real roots. $\endgroup$ – md2perpe Jun 9 '17 at 17:36
  • $\begingroup$ How do you know $a>0$? $\endgroup$ – Pedro Gomes Jun 9 '17 at 17:37
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Consider the parabola $y=ax^2+bx+c$. If the right hand side has distinct real roots, the vertex will have negative $y$-coordinate (as $a=\langle u,u\rangle>0$).

So the only case when no point on the parabola has negative $y$-coordinate is when $b^2-4ac\le0$.

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If $at^2+bt+c \geq 0$ then subbing $t=-\frac{b}{2a}$ you get $$a(\frac{-b}{2a})^2+b(-\frac{b}{2a})+c \geq 0$$

Now, since $a>0$, multiplying by $4a$ you get $$b^2-2b^2+4ac \geq 0$$

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