Solution of problem S596

Let a, b, c be the side–lengths of a triangle. Prove that

\frac{a}{b + c − a} + \frac{b}{c + a − b} + \frac{a}{b + c − a} ≥ \frac{3(a^{2} + b^{2} + c^{2})}{ab + bc + ca}

Proof Let’s change a + b − c = z, b + c − a = x, c + a − b = y, yielding a = (y+z)/2, b = (a+b)/2, c = (x+y)/2. The inequality becomes equivalent to

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

and this is simply AGM.

3(x^{3}y^{2} + x^{3}z^{2}) ≥ 6x^{3}yz and cyclic

yield

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

x^{3}y^{2} + x^{3}y^{2} + z^{3}y^{2} ≥ 3x^{2}y^{2}z and cyclic x^{4}z + y^{4}z ≥ 2x^{2}y^{2}z and cyclic

complete the proof.