Proposed solution of problem O427

Dear Editor, I would like to submit the following solution of problem O427

Let ABC be a triangle and m_{a}, m_{b}, m_{c} be the lengths of its medians. Prove that

\sqrt{3}(am_{a} + bm_{b} + cm_{c}) ≤ 2s^{2}

Proof The inequality is

\sqrt{3}(a + b + c) ∑_{cyc} \frac{a}{a + b + c)}\sqrt{\frac{2b^{2} + 2c^{2} − a^{2}}{4}} ≤ \frac{(a + b + c)^{2}}{2}

The concavity of \sqrt{x} yields

\sqrt{3}(a+b+c) ∑_{cyc} \frac{a}{a + b + c)}\sqrt{\frac{2b^{2} + 2c^{2} − a^{2}}{4}} ≤ \sqrt{3}(a+b+c)\sqrt{∑_{cyc} \frac{a}{a + b + c)} \frac{2b^{2} + 2c^{2} − a^{2}}{4}}

so it suffices to prove

\sqrt{3}(a + b + c)\sqrt{∑_{cyc} \frac{a}{a + b + c)} \frac{2b^{2} + 2c^{2} − a^{2}}{4}} ≤ \frac{(a + b + c)^{2}}{2}

or squaring

3 ∑_{cyc} a(2b^{2} + 2c^{2} − a^{2}) ≤ (a + b + c)^{3}

Now let’s set a = y + z, b = x + z, c = x + y, x, y, z ≥ 0 and the inequality becomes

∑_{sym}(x^{3} + 3x^{2}y) ≥ 24xyz

and this is clearly AGM.

Roma 10/14/2017 Best regards Paolo Perfetti

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