Problem 12136
(American Mathematical Monthly, Vol.126, October 2019) Proposed by A. Stadler (Switzerland).
Prove
a2+ b2+ c2≥ a4
rb4+ c4 2 + b4
rc4+ a4 2 + c4
ra4+ b4 2 for all positive real numbersa, b, and c.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Since 2(x2− xy + y2)2− (x4+ y4) = (x − y)4≥ 0, it follows that px2− xy + y2≥ 4
rx4+ y4 2 for all real x, y and it suffices to show the stronger inequality
a2+ b2+ c2≥ ap
b2− bc + c2+ bp
c2− ca + a2+ cp
a2− ab + b2. Moreover, the concavity of x → f(x) =√
x implies that
f a(b2− bc + c2) + b(c2− ca + a2) + c(a2− ab + b2) a + b + c
≥af (b2− bc + c2) + bf (c2− ca + a2) + cf (a2− ab + b2) a + b + c
that is
√a + b + c s
X
sym
ab2− 3abc ≥ ap
b2− bc + c2+ bp
c2− ca + a2+ cp
a2− ab + b2
Hence it remains to show that
(a2+ b2+ c2)2≥ (a + b + c) X
sym
ab2− 3abc
!
that is
a2(a − b)(a − c) + b2(b − a)(b − c) + c2(c − a)(c − b) ≥ 0
which holds by Schur’s inequality.
Remark: the given inequality holds for all real numbers a, b, c. It does not hold if we replace
4
q(·)4+(·)4
2 with n
q(·)n+(·)n
2 with any integer n ≥ 5. Take a = 1/2, b = c = 1.
