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?

  • 3
    This implies that $at^{2}+bt+c \geq 0$ For all $t$. The universal quantifier is important. – Brian Borchers Sep 26 at 15:09
up vote 6 down vote accepted

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 at 13:50
  • Thank you so much sir, for adding details. – Akash Patalwanshi Sep 26 at 13:59
  • 1
    @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 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?

  • 1
    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 at 13:53
  • Thank you so much sir – Akash Patalwanshi Sep 26 at 13:57
  • This is the most intuitive answer to the question. +1. – MPW Sep 26 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.

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