Page 14 - MATINF Nr. 9-10

14 P.H. Hoai

2
2
2
2 |xy| + 2 |yz| + 2 |zx| ≥ x + y + z .
x + y
By x + y + z + xyz = 0, there exist xy ≥ 0 and z = − .
1 + xy
We need to prove
2
2
2
2|xy| + 2|yz| + 2|zx| ≥ x + y + z , i.e.
2 (x + y) 2 Å x + y ã 2
2
2
2xy + ≥ x + y + , i.e.
1 + xy 1 + xy
4xy 2 1
+ ≥ 1 + .
(x + y) 2 1 + xy (1 + xy) 2
2
2
2
2
But (1 − x ) (1 − y ) ≥ 0, then (x + y) ≤ (1 + xy) , and hence
4xy 2 4xy 2 1
+ ≥ + ≥ 1 + ,
(x + y) 2 1 + xy (1 + xy) 2 1 + xy (1 + xy) 2
since 0 ≤ xy ≤ 4.

Å ã 2 Å ã 2 Å ã 2
b − c c − a a − b 16abc
+ + + ≥ 2. (3)
b + c c + a a + b (a + b) (b + c) (c + a)
This inequality follows immediately from the identity
2
2
Å ã 2 Å ã 2 Å ã 2 Å ã Å ã Å ã 2
b − c c − a a − b 16abc b − c c − a a − b
+ + + = 2 + .
b + c c + a a + b (a + b) (b + c) (c + a) b + c c + a a + b
We back to the given problem. Using Minkowski’s Inequality, (1), (2) and (3), we have

8a Å b − c ã 2 8b Å c − a ã 2 8c Å a − b ã 2
+ + + + +
b + c b + c c + a c + a a + b a + b
Ã
! 2 " # 2
… … Å ã 2 Å ã 2 Å ã 2
a b c b − c c − a a − b
≥ 8 + + + + +
b + c c + a a + b b + c c + a a + b
s
ñ ô
4abc b − c c − a a − b
Å ã Å ã 2 Å ã 2 Å ã 2
≥ 8 4 + + 2 + +
(a + b) (b + c) (c + a) b + c c + a a + b
s
ñ ô
Å ã 2 Å ã 2 Å ã 2
b − c c − a a − b 16abc
= 32 + 2 + + + ≥ 6.
b + c c + a a + b (a + b) (b + c) (c + a)

Problem 2. Let a, b, c three non-negative real numbers such that ab + bc + ca > 0, and let
1
k ≥ . Prove that:
4

2
k a bc k b ca k c ab
2
2
+ + + + +
b + c (b + c) 2 c + a (c + a) 2 a + b (a + b) 2
s
Å ã 2 2
1 4 (k − 2) abc
≥ 2k + + .
2 (a + b) (b + c) (c + a)