A Moscow Olympiad Question with Two Inequalities

Solution 1

Consider the quadratic polynomial $f(x)=ax^2+bx+c$ and observe that $f(0)=c$ while $f(1)=a+b+c.$ Thus, it is given that $f(0)f(1)\lt 0.$ It follows that $f$ has a root on $(0,1).$ This is only possible when the discriminant $b^2-4ac$ of the quadratic function is positive.

Solution 2

Suppose $b^2\le 4ac.$ Then

$\displaystyle (a+b+c)c=ac+bc+c^2 \ge \frac{b^2}{4}+bc+c^2=\left(\frac{b}{2}+c\right)^2\ge 0.$

This is not true. Thus, our supposition is incorrect and $b^2\gt 4ac.$

Acknowledgment

This is problem 1999-C1 Moscow Mathematical Olympiads, 1993-1999 by R. Fedorov, A. Belov, A. Kovaldzhi, I. Ivashchenko (MSRI/AMS, 2011).

Solution 2 is by Amit Itagi.