Let a, b, c, and d be positive real numbers

Problem 12115

(American Mathematical Monthly, Vol.126, May 2019) Proposed by M. Dr˘agan (Romania).

Let a, b, c, and d be positive real numbers. Prove

(a³+ b³)(a³+ c³)(a³+ d³)(b³+ c³)(b³+ d³)(c³+ d³) ≥ (a²b²c²+ a²b²d²+ a²c²d²+ b²c²d²)³.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We will show a more general inequality: if p ≥ 2 and a, b, c, d ≥ 0 then

(a^p+ b^p)(a^p+ c^p)(a^p+ d^p)(b^p+ c^p)(b^p+ d^p)(c^p+ d^p) ≥ 2^6−2p(a²b²c²+ a²b²d²+ a²c²d²+ b²c²d²)^p. By the Power Mean Inequality, if x, y ≥ 0 then x^p+ y^p≥2^1−p/2(x²+ y²)^p/2 and it follows that the left-hand side is greater or equal to

2^6−3p (a²+ b²)(a²+ c²)(a²+ d²)(b²+ c²)(b²+ d²)(c²+ d²)
p/2

. Hence it remains to show that

(a²+ b²)(a²+ c²)(a²+ d²)(b²+ c²)(b²+ d²)(c²+ d²) ≥ 4(a²b²c²+ a²b²d²+ a²c²d²+ b²c²d²)² that is the generalized inequality for p = 2.

After expanding both sides multiplied by 3, it reduces to 3X

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a⁶b⁴c²d⁰+X

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a⁶b²c²d²≥3X

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a⁴b⁴c²d²+X

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a⁴b⁴c⁴d⁰

which holds because, by Muirhead’s inequality, X

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a⁶b⁴c²d⁰≥X

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a⁴b⁴c²d² and X

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a⁶b²c²d²≥X

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a⁴b⁴c⁴d⁰.