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Algebra Difficulty 5.6 AIME, harder Prove it Singapore

Let a,b,ca, b, c be real numbers such that 0<a,b,c<1/20 < a, b, c < 1/2 and a+b+c=1a + b + c = 1. Prove that for all real numbers x,y,zx, y, z,
abc(x+y+z)2≥ayz(1−2a)+bxz(1−2b)+cxy(1−2c). abc(x + y + z)^2 \geq ayz(1 - 2a) + bxz(1 - 2b) + cxy(1 - 2c).
When does equality hold?

Solution

By symmetry, we may assume that xa≥yb≥zc\frac{x}{a} \ge \frac{y}{b} \ge \frac{z}{c}. Let yb=q\frac{y}{b} = q, xa=q+α\frac{x}{a} = q + \alpha and zc=q−β\frac{z}{c} = q - \beta where α,β≥0\alpha, \beta \ge 0. Thus x=a(q+α)x = a(q + \alpha), y=bqy = bq, z=c(q−β)z = c(q - \beta). We have
abc(x+y+z)2≥ayz(1−2a)+bxz(1−2b)+cxy(1−2c)⇔ (aq+aα+bq+cq−cβ)2≥q(q+α)(1−2c)+q(q−β)(1−2a)q(q+α)(1−2c)++(q−β)(q+α)(1−2b)⇔ [q+(aα−cβ)]2≥q2+2q(aα−cβ)−αβ(1−2b)⇔ (aα−cβ)2≥−αβ(1−2b) \begin{align*} & abc(x + y + z)^2 \ge ayz(1 - 2a) + bxz(1 - 2b) + cxy(1 - 2c) \\ \Leftrightarrow\ & (aq + a\alpha + bq + cq - c\beta)^2 \ge q(q + \alpha)(1 - 2c) + q(q - \beta)(1 - 2a) \\ & \phantom{q(q + \alpha)(1 - 2c) +} + (q - \beta)(q + \alpha)(1 - 2b) \\ \Leftrightarrow\ & [q + (a\alpha - c\beta)]^2 \ge q^2 + 2q(a\alpha - c\beta) - \alpha\beta(1 - 2b) \\ \Leftrightarrow\ & (a\alpha - c\beta)^2 \ge -\alpha\beta(1 - 2b) \end{align*}
Since α,β≥0\alpha, \beta \ge 0 and b≤0.5b \le 0.5, the last inequality holds. Also equality holds iff x/a=y/b=z/cx/a = y/b = z/c.

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