The following simple exercise appears in the book "Undergraduate Topology" by Robert H. Kasriel (Dover publication 2009). I know you can find proofs in any elementary algebra book, yet I made the effort proving this trying to use the most elementary algebra and made for the fun two version of it, the first very straightforward, the second (unnecessary lengthy) making me ring alarm bells:
To prove: Consider the function $f$ given by $f (x) = A x^2+ 2Bx+ C$, where $A>0$, and which further satisfies $f(x) \geq 0$ for all real $x$. Prove that $B^2-AC\leq 0$.
Proof 1: $f(x)\geq 0,\, \forall x\in \mathbb{R}$, then $f(-\frac{B}{A})\geq 0$ which gives
\begin{align}f(-\frac{B}{A})&= \frac{B^2}{A}-2\frac{B^2}{A}+C \geq 0\\ \Rightarrow\quad &B^2-AC \leq 0 \end{align}
Proof 2 (unnecessary lengthy and probably with a dubious step):
\begin{align} f (x) &= A x^2+ 2Bx+ C\\ &= A\left(x^2+ 2\frac{B}{A}x+ \frac{C}{A}\right)\quad \text{ (as } A>0)\\ &= A\left[x^2+ \left(\frac{B}{A}+\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }+\frac{B}{A}-\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)x+ \left(\frac{B}{A}+\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)\left(\frac{B}{A}-\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)\right]\quad (*)\\ &= A\left[x+ \left(\frac{B}{A}+\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)\right]\left[x+ \left(\frac{B}{A}-\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)\right] \end{align} We know that $f(x)\geq0,\, \forall x\in \mathbb{R}$. So $f(x)\geq 0 $ for $x=-\frac{B}{A}$ and get \begin{align*} 0&\leq A\left(-\frac{B}{A}+ \left(\frac{B}{A}+\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)\right)\left(-\frac{B}{A}+ \left(\frac{B}{A}-\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\right)\right)\\ &= -A\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\sqrt{\frac{B^2}{A^2}-\frac{C}{A} }\\ &= A\left(\frac{C}{A}-\frac{B^2}{A^2}\right) \\ \end{align*} which requires $ \frac{C}{A}-\frac{B^2}{A^2}\geq 0$ or $ B^2-AC\leq 0$ $$\blacklozenge$$ My doubts about the soundness of the logic is at line (*), where I introduce a quantity which I want to prove that it is negative, yet take the square root of it. My conclusion is that this proof is not acceptable but perhaps someone has good arguments to find it OK (not elegant but OK)