IMO 2001 · Problem 2

Prove that \[ \frac{a}{\sqrt{a^2+8bc}}+ \frac{b}{\sqrt{b^2+8ca}}+ \frac{c}{\sqrt{c^2+8ab}}\geq 1 \] for all positive real numbers \(a\), \(b\) and \(c\).

Official text of the problem. Official problem paper (PDF)

Solution

Put \(s=a+b+c\). By the arithmetic–geometric mean inequality, \[ a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\geq 6abc. \] Consequently, \[ \begin{aligned} s^3-(a^3+b^3+c^3+24abc) &=3\bigl(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b-6abc\bigr)\\ &\geq 0. \end{aligned} \]

Now set \[ T=a\sqrt{a^2+8bc}+b\sqrt{b^2+8ca}+c\sqrt{c^2+8ab}. \] Applying the Cauchy–Schwarz inequality gives \[ \begin{aligned} T^2 &\leq (a+b+c)\bigl(a(a^2+8bc)+b(b^2+8ca)+c(c^2+8ab)\bigr)\\ &=s(a^3+b^3+c^3+24abc)\\ &\leq s^4. \end{aligned} \] Thus \(T\leq s^2\).

A second application of Cauchy–Schwarz yields \[ \left( \frac{a}{\sqrt{a^2+8bc}}+ \frac{b}{\sqrt{b^2+8ca}}+ \frac{c}{\sqrt{c^2+8ab}} \right)T \geq (a+b+c)^2=s^2. \] Since \(T>0\), it follows that \[ \frac{a}{\sqrt{a^2+8bc}}+ \frac{b}{\sqrt{b^2+8ca}}+ \frac{c}{\sqrt{c^2+8ab}} \geq \frac{s^2}{T}\geq 1, \] as required.

Equality holds if and only if \(a=b=c\): this condition makes every term equal to \(1/3\), while equality in the proof requires equality in the six-term arithmetic–geometric mean inequality, which forces \(a=b=c\).

The solution was checked against Evan Chen's published IMO solution notes.