MathDB
Prove this 6 variable inequality

Source: 2019 Jozsef Wildt International Math Competition

May 20, 2020
Summationinequalities

Problem Statement

If bkak0b_k \geq a_k \geq 0 (k=1,2,3)(k = 1, 2, 3) and α1\alpha \geq 1 then(α+3)cyc(b1a1)((b2+b3)α+2+(a2+a3)α+2(a2+b3)α+1(b2+a3)α+1)(\alpha+3)\sum \limits_{cyc}(b_1-a_1)\left((b_2+b_3)^{\alpha+2}+(a_2+a_3)^{\alpha+2}-(a_2+b_3)^{\alpha+1}-(b_2+a_3)^{\alpha+1}\right) (α+2)(α+3)cyc(b1a1)(b2a2)(b3α+1a3α+1)\leq (\alpha+2)(\alpha+3)\sum \limits_{cyc}(b_1-a_1)(b_2-a_2)(b_3^{\alpha+1}-a_3^{\alpha+1}) +(b3+b2+a1)α+3+(b3+a2+a1)α+3+(a3+b2+a1)α+3+(a3+a2+b1)α+3+ (b_3 + b_2 + a_1)^{\alpha+3}+(b_3 + a_2 + a_1)^{\alpha+3}+(a_3 + b_2 + a_1)^{\alpha+3}+(a_3 + a_2 + b_1)^{\alpha+3} (b3+b2+b1)α+3(b3+a2+a1)α+3(a3+b2+b1)α+3(a3+a2+a1)α+3-(b_3 + b_2 + b_1)^{\alpha+3}-(b_3 + a_2 + a_1)^{\alpha+3}-(a_3 + b_2 + b_1)^{\alpha+3}-(a_3 + a_2 + a_1)^{\alpha+3}