Proposed solution of problem 4246

Dear Editor of Crux Mathematicorum, I would like to submit the following solution of problem 4246

Find the lowest lower bound for abc + bcd + cda + dab over all positive a, b, c and d satisfying

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

Proof Let’s suppose b = c = d. The condition

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

yields a = \frac{3 − 3b^{2} + \sqrt{9 − 14b^{2} + 9b^{4}}}{2b} ∼ \frac{3}{b} as b → 0. It follows that

abc + bcd + cda + dab ∼ 3b + b^{3} + 3b + 3b = 9b + b^{3}

so the infimum (not minimum) is zero

Roma, (Italy) 03/19/2017 Thank you, Best regards Paolo Perfetti

Perfetti Paolo, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, 00133 Roma, Italy – email: perfetti@mat.uniroma2.it