Why is the discriminant of quadratic $f$, $\Delta \leq 0$, when $f(x) \geq 0 \forall x \in \mathbb{R}$?

Question

I was reading a proof of the Cauchy-Schwarz inequality (elementary form). However, I was unable to understand some part of it. Here is the proof:

Proof: Consider the following quadratic polynomial $$f(x)=\sum_ {i=1}^ n (a_ix-b_i)^2=(\sum_ {i=1}^ n a_i^2)x^2-2(\sum_ {i=1}^ n a_ib_i)x+\sum_ {i=1}^ n b_i^2$$ $\color{red}{\textrm{Since $f(x) \geq 0$ for any $x \in \mathbb{R}$, it follows that the discriminant of $f(x)$ is negative or zero}}$, i.e., $$4(\sum_ {i=1}^ n a_ib_i)^2-4(\sum_ {i=1}^ n a_i^2)(\sum_ {i=1}^ n b_i^2) \leq 0$$ This follows the desired inequality. $\blacksquare$

---

I am unable to understand the red part. I understand that $f(x) \geq 0$ for any $x \in \mathbb{R}$. But how does this imply that the discriminant is negative or zero?

Answer 1

See

$f(x)=\sum_ {i=1}^ n (a_ix-b_i)^2$

Clearly $f(x)$ will be non-negative as it is sum of squares of real some real numbers

Therefore we can write $f(x)\geq 0$

Now let's turn to otgere form of $f(x)$

$f(x)=(\sum_ {i=1}^ n a_i^2)x^2-2(\sum_ {i=1}^ n a_ib_i)x+\sum_ {i=1}^ n b_i^2$

Clearly this is a quadratic equation (assuming at least when $a_i\neq 0$)

Now coefficient of $x^2$ is $\sum_ {i=1}^ n a_i^2$ which will be always positive (under our assumption), therefore the parabola corresponding to equation will open upwards.

Three cases arise:

$1)$ $D<0$

In this case our quadratic equation will have no real roots , therefore to satisfy this condition our parabola must always be above $x-$ axis because if it comes a little down the x-axis then it will have to cross it ( since it will extend to $\infty$

) giving two real roots. Therefore parabola remains above x-axis and hence y-coordinate of all points on parabola is positive. Hence $D<0$ satisfies $f(x)>0$ for all real x

$2)$ $D=0$

In this case our quadratic equation will have repeated roots therefore the parabola will only touch x-axis while remaining above it only. therefore this condition satisfies $f(x)\geq 0$ for all real x

$3)$ $D>0$

Now the equation will have two real roots and therefore parabola will be below $x-$ axis between the roots therefore $f(x)\ngeq 0$ for all real x

Therefore now you can easily figure out what does red line wants to convey.

Answer 2

Let $P(x)=ax^2+bx+c$ and suppose $P(x)\geq 0$ for all $x$.

• Case 1: $a=0$. In this case, $P(x)=bx+c$ and $P(x)\geq 0$ for all $x$, so it follows $b=0$ (think of what the graph of a straight line looks like if $b\neq 0$). So, $b^2-4ac=0$.

• Case 2: $a> 0$. Then, we can complete the square, $P(x)=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}$ is $\geq 0$ for all $x$, so in particular when $x=-\frac{b}{2a}$, and thus $c-\frac{b^2}{4a}\geq 0$. Rearranging yields $b^2-4ac\leq 0$ (where did I use the fact $a>0$?).

This completes the proof (Why is case 3 of $a<0$ not possible?)

Answer 3

If $D<0$, this means that the quadratic has no real roots. This forces us to conclude that the graph of the function must either lie completely above, or completely below the $x$-axis, since it can neither touch nor intersect it. Since the coefficient of $x^2$ is positive, this means that the parabola must be upward-opening (because any curve of this nature is of the form $a(x+b)^2+c$, hence can be formed by shifting and compressing/expanding the graph of $y=x^2$). So it can't lie below the $x$-axis, hence it must lie above.