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Phika sextuplets satisfy a strange inequality

Source: 2022 3rd OMpD L2 P4 / L3 P2 - Brazil - Olimpíada Matemáticos por Diversão

July 8, 2023
algebrainequalitiespositive real

Problem Statement

We say that a sextuple of positive real numbers (a1,a2,a3,b1,b2,b3)(a_1, a_2, a_3, b_1, b_2, b_3) is <spanclass=′latex−italic′>phika</span><span class='latex-italic'>phika</span> if a1+a2+a3=b1+b2+b3=1a_1 + a_2 + a_3 = b_1 + b_2 + b_3 = 1.
(a) Prove that there exists a <spanclass=′latex−italic′>phika</span><span class='latex-italic'>phika</span> sextuple (a1,a2,a3,b1,b2,b3)(a_1, a_2, a_3, b_1, b_2, b_3) such that: a1(b1+a2)+a2(b2+a3)+a3(b3+a1)>1−120222022a_1(\sqrt{b_1} + a_2) + a_2(\sqrt{b_2} + a_3) + a_3(\sqrt{b_3} + a_1) > 1 - \frac{1}{2022^{2022}}
(b) Prove that for every <spanclass=′latex−italic′>phika</span><span class='latex-italic'>phika</span> sextuple (a1,a2,a3,b1,b2,b3)(a_1, a_2, a_3, b_1, b_2, b_3), we have: a1(b1+a2)+a2(b2+a3)+a3(b3+a1)<1a_1(\sqrt{b_1} + a_2) + a_2(\sqrt{b_2} + a_3) + a_3(\sqrt{b_3} + a_1) < 1
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