Proposed solution to problem J38 – deadline 03–15–07

Perfetti Paolo, dipartimento di matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, 00133 Roma, Italy – email: perfetti@mat.uniroma2.it

Dear Editor, I would like to submit the following solution of problem J38 (deadline 03– 15–07)

Let a, b, c positive real numbers. Prove that

\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} ≥ \frac{a^{2} + bc}{(a + b)(a + c)} + \frac{b^{2} + ac}{(b + a)(b + c)} + \frac{c^{2} + ab}{(c + a)(c + b)}

The inequality is equivalent to

\frac{a^{3} + b^{3} + c^{3} + 3abc − ab(a + b) − ac(a + c) − bc(b + c)}{(a + b)(a + c)(b + c)} ≥ 0

which is in turn equivalent to

a(a − b)(a − c) + b(b − a)(b − c) + c(c − a)(c − b) ≥ 0

This is the well–known Schur’s inequality which is true of course and whose proof we write just for completeness. By symmetry we can suppose a ≥ b ≥ c hence

\begin{matrix}a(a − b)(a − c) + b(b − a)(b − c) + c(c − a)(c − b) ≥ \\ a(a − b)(b − c) − b(a − b)(b − c) + c(a − c)(b − c) = (a − b)^{2}(b − c) + c(a − c)(b − c) ≥ 0 \end{matrix}

Best regards Paolo Perfetti