Proposed solution to problem 1038

Let m be a nonnegative real number and x, y be positive real numbers. Prove that for any triangle ABC with side lengths a, b, c where [ABC] denotes the area of the triangle,

\frac{a^{m+2}}{(xb + yc)^{m}} + \frac{b^{m+2}}{(xc + ya)^{m}} + \frac{c^{m+2}}{(xa + yb)^{m}} ≥ \frac{4\sqrt{3}}{(x + y)^{m}} · [ABC]

Proof By Hölder’s inequality

∑_{cyc(a,b,c)} \frac{a^{m+2}}{(xb + yc)^{m}} ≥ \frac{(a + b + c)^{m+2}}{(xb + yc + xc + ya + xa + yb)^{m}} 3^{1+m−m−2}

so it suffices to show that

\frac{(a + b + c)^{m+2}}{3(x + y)^{m}(a + b + c)^{m}} ≥ \frac{4\sqrt{3}}{(x + y)^{m}} · [ABC]

that is

(a + b + c)^{2} ≥ 12\sqrt{3}[ABC] = 12\sqrt{3}\sqrt{\frac{a + b + c}{2} \frac{a + b − c}{2} \frac{a − b + c}{2} \frac{−a + b + c}{2}}

With x = (b + c − a)/2 y = (c + a − b)/2 z = (a + b − c)/2 this last is equivalent to

4(x + y + z)^{2} ≥ 12\sqrt{3}\sqrt{(x + y + z)xyz} ⟺ (x + y + x)^{\frac{3}{2}} ≥ 3\sqrt{3}\sqrt{xyz} that is

(x + y + z)^{3} ≥ 27(xyz)