Page 17 - MATINF Nr. 9-10

Three beautiful-difficult inequality problems 17

Case B. 3 ≤ u ≤ 4, so

4 (u + 6) 3 4 (4 + 6) 3 Å 1 ã 1
2
0
0
f (v) ≥ f (4 − u) = 2 + 2 k − 2 ≥ 2 + 2 − 2 = > 0,
(5u + 12) (5 × 4 + 12) 16 32
and hence
2
(12 − u) (u + 6)
2
2
2
f (v) ≥ f (4 − u) = + 2 k − 1 u + 2 k − 2 (4 − u) − 8k − 17
5u + 12
5 (4 − u) (u − 3) 2
= ≥ 0.
5u + 12
The proof is complete.

Remark 1. The above result can be rewritten in the following form

2
2
2
a k bc b k ca c k ab
+ + + + +
b + c (b + c) 2 c + a (c + a) 2 a + b (a + b) 2
s
k 4 (2k − 1) abc
Å ã 2 2
≥ + 2 + ,
2 (a + b) (b + c) (c + a)
where 0 ≤ k ≤ 4.
Problem 3. Let a, b, c three non-negative real numbers such that abc = 1, and let k ≥ −2.
Prove that:

3
2
2

2
2
2

3
a + b + c + k + 3k − 2 (ab + bc + ca) ≥ 3k + 3k (a + b + c) + 3k − 9k − 3.
Solution. We first prove the following inequality
1 1 ã
Å
3
2
2
2
2
a + b + k + 3k − 2 + − 3k + 3k (a + b)
a b
2 Ä √ ä
2
3
2
≥ 2ab + k + 3k − 2 √ . − 3k + 3k 2 ab , (6)
ab
√
Ä √ ä 2
a − b √
2 3 2 2 Ä √ ä 2
i.e. (a − b) + k + 3k − 2 − 3k + 3k a − b ≥ 0,
ab
√ ï √ 3 2 ò
Ä √ ä 2 Ä √ ä 2 k + 3k − 2
2
i.e. a − b a + b + − 3k + 3k ≥ 0.
ab
Indeed, we have
√ 3 2 √ 3 2
Ä √ ä 2 k + 3k − 2 k + 3k − 2 4
2
3
a + b + ≥ 4 ab + = k + 3k − 2 c + √ ,
ab ab c
and hence we need to prove that
4
2
3
2

k + 3k − 2 c + √ − 3k + 3k ≥ 0, i.e.
c
√ √
√ √ 2 ( c + 2) ( c − 1) 2
3 c − 1 (k − 1) (k + 2) + (k + 2) (k − 1) 2 c + √ ≥ 0.
c