bel.jad5
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x, y, z x 2 +y 2 +z 2 =3 x 4 y 4 + y 4 z 4 + z 4 x 4 +(xyz) 3 ≥ 4(xyz) 2 xyz = w 3 3= x 2 + y 2 + z 2 ≥ 3 3 p x 2 y 2 z 2 =3w 2 ⇒ w ≤ 1 x 4 y 4 + y 4 z 4 + z 4 x 4 +3 3 p x 8 y 8 z 8 ≥ 2x 2 y 2 z 2 (x 2 + y 2 + z 2 ) ⇔ x 4 y 4 + y 4 z 4 + z 4 x 4 +3 3 p x 8 y 8 z 8 ≥ 6x 2 y 2 z 2 6(xyz) 2 - 3(xyz) 8 3 ≥ 4(xyz) 2 - (xyz) 3 ⇔ (xyz) 3 - 3(xyz) 8 3 + 2(xyz) 2 ≥ 0 ⇔ xyz - 3(xyz) 2 3 +2 ≥ 0 ⇔ w 3 - 3w 2 +2 ≥ 0 ⇔ (w - 1)(w 2 - 2w - 2) ≥ 0 w - 1 ≤ 0 w 2 - 2w - 2 ≤ 0 w =1 w =0 (x, y, z) = (1, 1, 1) (x, y, z) = (0, 0, √ 3)
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Transcript of bel.jad5
Nermin Hodzic, July 2015
Let x, y, z be nonnegative reals such that x2+y2+z2 = 3.Prove the inequalityx4y4 + y4z4 + z4x4 + (xyz)3 ≥ 4(xyz)2
Solution;
Let xyz = w3.Using AmGm inequality we have
3 = x2 + y2 + z2 ≥ 3 3√x2y2z2 = 3w2 ⇒ w ≤ 1
Using Schleifer inequality we have
x4y4 + y4z4 + z4x4 + 3 3√x8y8z8 ≥ 2x2y2z2(x2 + y2 + z2)⇔
x4y4 + y4z4 + z4x4 + 3 3√x8y8z8 ≥ 6x2y2z2
Hence it is su�cies to prove
6(xyz)2 − 3(xyz)83 ≥ 4(xyz)2 − (xyz)3 ⇔
(xyz)3 − 3(xyz)83 + 2(xyz)2 ≥ 0⇔
xyz − 3(xyz)23 + 2 ≥ 0⇔
w3 − 3w2 + 2 ≥ 0⇔ (w − 1)(w2 − 2w − 2) ≥ 0
Which is obvious since w − 1 ≤ 0 and w2 − 2w − 2 ≤ 0Equality holds if and only if w = 1 or w = 0,which yields (x, y, z) = (1, 1, 1) or (x, y, z) = (0, 0,
√3) and its cyclic permuta-
tions.
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