# Understanding the proof of Cauchy-Schwartz inequality

While reading the proof of Cauchy-schwarz inequality, I didn't get one step. The step is as below,

by positivity axiom, for any real number $$t$$

$$0≤⟨tu+v,tu+v⟩=⟨u,u⟩t^2+ 2⟨u,v⟩t+⟨v,v⟩$$

This imply $$0≤at^2 + bt+c$$ where $$a=⟨u,u⟩$$, $$b=2⟨u,v⟩$$ and $$c=⟨v,v⟩$$

After this they had written, this inequality implies that the quadratic polynomial has either no real roots or repeated real roots!

I didn't get this! How the quadratic polynomial $$at^2 + bt+c$$ has either no real roots or repeated real root?

• This implies that $at^{2}+bt+c \geq 0$ For all $t$. The universal quantifier is important. – Brian Borchers Sep 26 '18 at 15:09

We need that $$at^2 + bt+c\ge 0,\, a\ge 0$$ and this is true if and only if

$$b^2-4ac \le 0$$

indeed recall that by quadratic formula

$$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

and the cases

• $$b^2-4ac=0$$ corresponds to repeated roots and the parabola $$y=at^2+bt+c$$ is tangent to $$x$$ axis

• $$b^2-4ac<0$$ corresponds to no real roots and the parabola $$y=at^2+bt+c$$ is above to $$x$$ axis

Refer also to the related: Quadratic Equation with imaginary roots.

• But how? Sir please elaborate – Akash Patalwanshi Sep 26 '18 at 13:50
• Thank you so much sir, for adding details. – Akash Patalwanshi Sep 26 '18 at 13:59
• @AkashPatalwanshi That's just a basic properties for quadratic equations and related parabola $y=ax^2+bx+c$. It is nice if you revise this topic. You are welcome! Bye – gimusi Sep 26 '18 at 14:01

You can even think in a geometric way. If $$at^2+bt+c\geq 0$$ for all $$t\in\mathbb{R}$$ then the parabola is never below the real axis. What does it tell us about the number of times it intersects the real axis?

• If you write $f(t)=at^2+bt+c$ then the intersections with the real axis are points where $f(t)=0$, not where $t=0$. So there can be at most one point where $f(t)=0$, which means at most one root. – Mark Sep 26 '18 at 13:53
• Thank you so much sir – Akash Patalwanshi Sep 26 '18 at 13:57
• This is the most intuitive answer to the question. +1. – MPW Sep 26 '18 at 18:00

If the quadratic polynomial $$at^2+bt+c$$ would have two real roots $$x_1,x_2$$, then it would have a one sign on $$(x_1,x_2)$$ and a different sign on $$(-\infty, x_1)\cup(x_2, \infty)$$.

Therefore, there would exist some set on which the polynomial is negative, which is a contradiction.

Since $$a = \langle u, u \rangle \geq 0$$, the parabola is concave-up. Its minimum is achieved at the vertex $$t = -b/2a$$, and the value of this minimum is $$a\left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c = \frac{b^2}{4a} - \frac{b^2}{2a}+c = c - \frac{b^2}{4a}$$ and since the whole parabola $$at^2 + bt + c \geq 0$$ for any value of $$t$$, this minimum must be at least zero, so $$c - \frac{b^2}{4a} \geq 0$$ Rearranging this inequality to $$4ac \geq b^2$$ and plugging back in the values of $$a, b, c$$ will give the Cauchy-Schwarz inequality.