Problem 12024
(American Mathematical Monthly, Vol.125, February 2018) Proposed by M. Cucoane¸s, M. Dr˘agan, and N. Stanciu (Romania).
Let x, y, and z be positive real numbers satisfying xyz= 1. Prove (x10+ y10+ z10)2≥3(x13+ y13+ z13).
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 the following stronger inequality: for any real numbers a, b, and c,
(a5+ b5+ c5)2≥3abc(a7+ b7+ c7). (1) Then, the given inequality follows by letting a = x2, b = y2and c = z2,
(x10+ y10+ z10)2≥3(xyz)2(x14+ y14+ z14) =3 2
X
sym
x16y2z2
≥3 2
X
sym
x46/3y7/3z7/3= 3(xyz)7/3(x13+ y13+ z13) = 3(x13+ y13+ z13)
where in the second line we applied the Muirhead’s inequality.
Proof of the inequality (1) (note that it has appeared as Monthly problem 11543 in 2010).
Without loss of generality we may assume that a ≥ b ≥ c. Let
Ta := 1
2(a7+ b7+ c7)(a + b + c) − b3c3(b + c)2, Tb:= 1
2(a7+ b7+ c7)(a + b + c) − c3a3(c + a)2, Tc:= 1
2(a7+ b7+ c7)(a + b + c) − a3b3(a + b)2. Then Ta≥Tb≥Tc and
Tb+ Tc= (a7+ b7+ c7)(a + b + c) − c3a3(c + a)2−a3b3(a + b)2
= a8+ b8+ c8+X
sym
a7b − c3a3(c + a)2−a3b3(a + b)2
≥a8+ b8+ c8+X
sym
a5b3−c3a3(c + a)2−a3b3(a + b)2
= (a5+ b5+ c5)(a3+ b3+ c3) − c3a3(c + a)2−a3b3(a + b)2
= (a4−(b4+ c4))2+ b3c3(b − c)2≥0 where in the third line we applied again the Muirhead’s inequality.
Hence Ta≥Tb≥0, (a − c)2≥(a − b)2, and finally it is straightforward to verify that (a5+ b5+ c5)2−3abc(a7+ b7+ c7) = (b − c)2Ta+ (a − c)2Tb+ (a − b)2Tc
≥(b − c)2Ta+ (a − b)2(Tb+ Tc) ≥ 0.