Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

Find the roots $r_{1}, r_{2}, r_{3}, r_{4}$ of the equation $4x^{4} - a x^{3} + b x^{2} - c x + 5 = 0$ , given that they are positive reals satisfying $r_{1}/2 + r_{2}/4 + r_{3}/5 + r_{4}/8 = 1$ .

Solution

Solution:

We have $r_{1} r_{2} r_{3} r_{4} = 5/4$ and hence $(r_{1}/2)(r_{2}/4)(r_{3}/5)(r_{4}/8) = 1/4^{4}$ . But AM/GM gives that $(r_{1}/2)(r_{2}/4)(r_{3}/5)(r_{4}/8) \leq \left( (r_{1}/2 + r_{2}/4 + r_{3}/5 + r_{4}/8)/4 \right)^{4} = 1/4^{4}$ with equality iff $r_{1}/2 = r_{2}/4 = r_{3}/5 = r_{4}/8$ . Hence we must have $r_{1} = 1/2$ , $r_{2} = 1$ , $r_{3} = 5/4$ , $r_{4} = 2$ .

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