Proposed solution of problem 5380

Dear Ted, I would like to submit the following solution to problem 5380

Let Δ(x, y, z) = 2(xy +yz +zx)−(x^{2} +y^{2} +z^{2}) and a, b, c be the side–lengths of a triangle ABC. Prove that

F^{2} ≥ \frac{3}{16} \frac{Δ(a^{3}, b^{3}, c^{3})}{Δ(a, b, c)}

where F is the area of ABC.

Proof As well known F^{2} = s(s − a)(s − b)(s − c) and s = (a + b + c)/2. Upon setting a = y + z, b = x + z, c = x + y, the inequality becomes

∑_{sym} (27x^{4}y^{2} + 21(xy)^{3} + 5(xyz)^{2}) ≥ ∑_{sym} (27x^{4}yz + 26x^{3}y^{2}z)

The third degree Schür inequality is

(a^{3} + b^{3} + c^{3}) + 3abc ≥ ∑_{sym} a^{2}b

which applied to the triple (xy), (yz), (zx), yields

5 ∑_{sym}(xy)^{3} + 5 ∑_{sym}(xyz)^{2} ≥ 10 ∑_{sym} x^{3}y^{2}z

The inequality becomes

∑_{sym} (27x^{4}y^{2} + 16(xy)^{3}) ≥ ∑_{sym} (27x^{4}yz + 16x^{3}y^{2}z)

and the proof is complete upon observing that by the AGM we have

x^{4}y^{2} + x^{4}z^{2} ≥ 2x^{4}yz, (xy)^{3} + (xy)^{3} + (xz)^{3} ≥ 3x^{3}y^{2}z

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

Roma(Italy) 02/18/2016 Best regards Paolo Perfetti