Page 18 - MATINF Nr. 9-10

18 P.H. Hoai

Case A. −2 ≤ k ≤ 1. We assume that c = min {a, b, c}, so c ∈ (0; 1], and hence
√
( c − 1) (k − 1) ≥ 0.
√
Case B. k ≥ 1. We assume that c = max {a, b, c}, so c ≥ 1, and hence ( c − 1) (k − 1) ≥ 0.
Through the above two cases, we see that (6) is true.

We back to the given problem. Using (6) we have

Å ã Å ã
2 1 √ 2
2
3
3
2
2
LHS − RHS ≥ + c + k + 3k − 2 + 2 c − 3k + 3k √ + c − 3k + 9k + 3,
c c c
and hence
√ 2 √ 2 √
(1 − c) (k − c) [c + 2 (k + 1) c + k + 3]
LHS − RHS ≥ ≥ 0,
c
since c > 0 and k ≥ −2.