Solution of problem J585

Let a, b, c > 1 be real numbers such that

\frac{1}{a − 1} + \frac{1}{b − 1} + \frac{1}{c − 1} = 1

Prove that

abc + 44 ≥ 9(a + b + c)

Proof By writing x = a − 1 > 0, y = b − 1 > 0, z = c − 1 > 0 the inequality becomes

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 ⟹ (x + 1)(y + 1)(z + 1) + 44 ≥ 9(x + y + z + 3)

that is

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 ⟹ (x + 1)(y + 1)(z + 1) + 44 ≥ 9(x + y + z + 3)

Now let’s set X = 1/x, Y = 1/y, Z = 1/z and we get

X + Y + Z = 1 ⟹ (\frac{1}{X} + 1)(\frac{1}{Y} + 1)(\frac{1}{Z} + 1) + 44 ≥ 9(\frac{1}{X} + \frac{1}{Y} + \frac{1}{Z} + 3)

namely

\frac{1}{XY Z} + ∑_{cyc} \frac{1}{X} + ∑_{cyc} \frac{1}{XY} + 1 + 44 ≥ 9 ∑_{cyc} \frac{1}{X} + 27

and upon simplifications we come to

\frac{1}{XY Z} + ∑_{cyc} \frac{1}{XY} + 18 ≥ 8 ∑_{cyc} \frac{1}{X}

\frac{1}{XY Z} +\frac{X + Y + Z}{XY Z} +18 ≥ 8\frac{XY + Y Z + ZX}{XY Z} ⟺ 2+18XY Z ≥ 8 ∑_{cyc} XY

(1)

This is a special version of Schür’s inequality (a, b, c > 0) a^{3}+b^{3}+c^{3}+3abc ≥ a^{2}b+b^{2}a+b^{2}c+c^{2}b+a^{2}c+c^{2}a ≥ (a+b+c)(ab+bc+ca)−3abc or (a+b+c)^{3}−3(a^{2}b+b^{2}a+b^{2}c+c^{2}b+a^{2}c+c^{2}a)−3abc ≥ (a+b+c)(ab+bc+ca)−3abc (a+b+c)^{3}−3((a+b+c)(ab+bc+ca)−3abc)−3abc ≥ (a+b+c)(ab+bc+ca)−3abc If a + b + c = 1 it becomes 1 − 3(ab + bc + ca) + 9abc ≥ ab + bc + ca ⟺ 1 + 9abc ≥ 4(ab + bc + ca) which is (1) and the proof is concluded.