Page 16 - MATINF Nr. 9-10

16 P.H. Hoai

and, by Minkowski’s Inequality,

Ç … å 2
2 Ä√
X » X ä 2
2
(a + b) (a + c) = a + ab + bc + ca
2
≥ (a + b + c) + 9 (ab + bc + ca) = x + 11y,
and hence we have

2
1 ï y Å X a ã ò 2 (k − 1) r (x + 11y)
LHS ≥ √ kp + 3 + − p + ,
y p b + c M
ô
ñ √ Å ã 2 ï Å ã ò
(k − 1) p y X a x x r 8r
2
LHS ≥ √ + 3 + + (k − 1) 2 + + 3 + ,
y p b + c y y M M
√ X a 1 X a 8r
ï Å ã ò 2 Å ã
2
LHS ≥ (k − 1) t + 2 + 3 + √ + (k − 1) 2 + ,
b + c t + 2 b + c M
h r i 1 Å X a ã 2 X 2a
2
LHS ≥ (k − 1) 2 t + (t + 3) + 3 + + (k − 1) +
M t + 2 b + c b + c
2
(k − 1) 8r
2
+2 (k − 1) + ,
M
a 1 Å a ã 2
î 2 ó X X
2
LHS ≥ (k − 1) + 2 (k − 1) + 3 + +
b + c t + 2 b + c
2
(k − 1) 8abc
2
+2 (k − 1) + ,
(a + b) (b + c) (c + a)
2
(k − 1) u v + 8 u 2
2
2
2
LHS ≥ + 3 + + 2 (k − 1) + (k − 1) v.
2 4u − v + 16 2
We need to prove that
2
2
(k − 1) u v + 8 u 2 2 2 Å 1 ã 2 (k − 2) v
+ 3 + + 2 (k − 1) + (k − 1) v ≥ 2k + + , i.e.
2 4u − v + 16 2 2 2
2
2
(k − 1) u v + 8 u 2 2 2 Å 1 ã 2 (k − 2) v
+ 3 + + 2 (k − 1) + (k − 1) v − 2k + − ≥ 0,
2 4u − v + 16 2 2 2
(v + 8) (u + 6) 2
2
2
2
i.e. f (v) = + 2 k − 1 u + 2 k − 2 v − 8k − 17 ≥ 0.
4u − v + 16
4 (u + 6) 3
0
2
We have f (v) = 2 + 2 (k − 2).
(4u − v + 16)
Case A. u > 4, so
3 3 Å ã
4 (u + 6) 4 (4 + 6) 1 1
0
0
2
f (v) ≥ f (0) = + 2 k − 2 ≥ + 2 − 2 = > 0,
(4u + 16) 2 (4 × 4 + 16) 2 16 32
and hence
2
8 (u + 6)
2
2
f (v) ≥ f (0) = + 2 k − 1 u − 8k − 17
4u + 16
ï Å 1 ã u − 4 ò
2
= (u − 4) 2 k − + ≥ 0.
16 8 (u + 4)