Page 15 - MATINF Nr. 9-10

Three beautiful-difficult inequality problems 15

Solution. Let
2
2
2
p = a + b + c, x = a + b + c , y = ab + bc + ca,
x
t = , r = abc, M = (a + b) (b + c) (c + a) .
y
By Schur’s Inequality we have

P
a 4abc a (a − b) (a − c)
X
+ − 2 = ≥ 0, so
b + c (a + b) (b + c) (c + a) (a + b) (b + c) (c + a)
8abc X 2a
≥ 4 − . (4)
(a + b) (b + c) (c + a) b + c
Also,

2
2
2
2
X a a + b + c 2 Å a + b + c 2 ã abc r
= + + 3 = t + (t + 3) . (5)
b + c ab + bc + ca ab + bc + ca (a + b) (b + c) (c + a) M
Let
2a 2b 2c 8abc
u = + + , v = .
b + c c + a a + b (a + b) (b + c) (c + a)
By (4) we have v ≥ max {0, 4 − u}.

4u − 3v
By (5) we have t = .
8 + v
Let
2
1 X Å bc ã 2 (k − 1) abc
LHS = √ ka + +
ab + bc + ca b + c b + c

and

Å ã 2 2
1 (k − 2) v
RHS = 2k + + .
2 2
By Minkowski’s Inequality we have

LHS ≥
s
2
1 ï X bc ò 2 (k − 1) abc h X » i 2
√ k (a + b + c) + + (a + b) (a + c) .
ab + bc + ca b + c (a + b) (b + c) (c + a)

But
Å ã
X bc X bc
= a + − (a + b + c)
b + c b + c
ab + bc + ca X a + b + c
= − (a + b + c)
a + b + c b + c
Å ã
ab + bc + ca X a
= 3 + − (a + b + c)
a + b + c b + c
y Å X a ã
= 3 + − p,
p b + c