# Proof of Cauchy-Schwarz inequality, why does this work?

My books says to prove that the following inequality is true, and to use it to prove Cauchy-Schwarz: $$(a_1x+b_1)^2+(a_2x+b_2)^2+(a_3x+b_3)^2+\dots+(a_nx+b_n)^2 \ge 0$$

This is easy to prove because by the trivial inequality each term on the LHS is $\ge 0$. However, to prove Cauchy-Schwarz using this this book gives the hint:

Write the left side as a quadratic equation in $x$ and note that a quadratic equation is non negative for all $x$ if and only if the discriminant is non positive.

If you multiply this out and set the discriminant $\le 0$, the Cauchy-Schwarz inequality follows very straightforwardly.

My question is: If there were no restrictions on $a$, $b$, or $x$ before I multiplied it out, why are there restrictions on them now?

I know that before I was considering a series of quadratics and now I'm considering one giant one, but still it's the same inequality as before.

• $(a-b)^2\ge 0$, so $a^2+b^2\ge 2ab$ (when $a,b\in\mathbb{R}$). There were no restrictions on $a$ or $b$, so why are there "restrictions" on them now? In fact, would you consider $a^2\ge 0$ to be a "restriction"? That holds with no "restrictions" on $a\in\mathbb{R}$. Here you are using the various properties of multiplication and addition, not just the properties of $a,b,x$. – wj32 Jun 17 '13 at 2:15

We know the LHS is a quadratic with non-positive discriminant, because it does not dip below the $x$-axis.
Here's a concrete example: $(2x-b)^2+(x-b)^2\ge 0$. This is a quadratic, namely $5x^2-6bx+2b^2$. The discriminant is $36b^2-40b^2=-4b^2$, which is non-positive, for all values of $x,b$. In this case it's easy to see that $-4b^2$ is non-positive, but if the quadratic were more complicated, the discriminant would necessarily be non-positive because the quadratic can be rearranged into the sum of squares.