Problem
ALG-B1-M06-P020 Bounding the product
Let \(a,b,c\) be real numbers such that \(a+b+c=0\) and \(a^2+b^2+c^2=2\). Prove that \[-\frac{2}{3\sqrt{3}}\le abc\le \frac{2}{3\sqrt{3}}.\]
View \(a,b,c\) as roots of a cubic polynomial and use the fact that it has three real roots.
From the conditions we get \(ab+bc+ca=-1\). Let \(q=abc\). Then \(a,b,c\) are three real roots of \(F(t)=t^3-t-q\).
If the cubic \(t^3-t-q\) has three real roots, then \(q\) must lie between the local maximum and local minimum values of the function \(t^3-t\). These extrema occur at \(3t^2-1=0\), so \(t=\pm\frac{1}{\sqrt{3}}\).
We get \(\left(\frac{1}{\sqrt{3}}\right)^3-\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}\) and \(\left(-\frac{1}{\sqrt{3}}\right)^3+\frac{1}{\sqrt{3}}=\frac{2}{3\sqrt{3}}\). Therefore \(-\frac{2}{3\sqrt{3}}\le q\le\frac{2}{3\sqrt{3}}\), as required.
A strong problem: one must not only form the polynomial, but also use the shape of its graph.