A question about the proof of Schwarz inequality

Question

There are many proofs of the Cauchy-Schwarz inequality, here's one of them: Consider the following quadratic polynomial: $$f(x)=\left(\sum_{i=1}^{n} a_i^2 \right)x^2-2\left(\sum_{i=1}^{n} a_ib_i \right)x+\sum_{i=1}^{n} b_i^2=\sum_{i=1}^{n} (a_ix-b_i)^2. $$ Since $f(x)\geqslant0$ for all $x\in\Bbb R$, it follows that the discriminant of $f(x)$ is negative, i.e., $$ \left(\sum_{i=1}^{n}a_ib_i\right)^2-\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)\leqslant0 $$ Therefore: $$\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)\geqslant\left(\sum_{i=1}^{n}a_ib_i\right)^2.$$

the question is: why are we able to say :"Since $f(x)\geqslant0$ for all $x\in\Bbb R$, it follows that the discriminant of $f(x)$ is negative" ??

Answer 1

The discriminant $D=B^2-4AC$ of a quadratic poinomial $Ax^2+Bx+C$ is the thing you take the square root of to find the two roots. Thus

• $D>0$ corresponds to two distinct real roots
• $D=0$ corresponds to a double real root
• $D<0$ corresponds to two distinct complex roots.

If $f(x)\ge 0$ for all $x$, there are certainly not two distinct real roots (between which it would have a different sign then outside). We conclude $D\le 0$.

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More down-to-earth you might notice that $$ f(x)=Ax^2+Bx+C=A\left(x+\frac B{2A}\right)^2-\frac{B^2-4AC}{4A}$$ and if $A>0$ this is $\ge -\frac{B^2-4AC}{4A}$ with equality at $x=-\frac B{2A}$, so we better have $ -\frac{B^2-4AC}{4A}\ge0$ if we want $f(x)\ge0$ for all $x$.

Answer 2

We are not able to say that the discriminant of $f(x)$ is negative. Look at exactly what you wrote:"Since $f(x) \ge 0$ for all x∈R, it follows that the discriminant of f(x) is negative, i.e., $$\left(\sum_{i=1}^{n}a_ib_i\right)^2-\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right)\leqslant0$$. This inequality does not follow from what you said (i.e. it is not a strict inequality, which your statement implies). We don't know the discriminant is negative, it could be $0$. We know it can't be positive, for then the quadratic would have two real roots, but since $f(x) \ge$ $0$ $\forall$ $x$ $ \in \mathbb{R}$, it can have at most 1 real root.