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ARTICOLE SI NOTE DE MATEMATICA
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Three beautiful-difficult inequality problems

Pham Huu Hoai 1

In this paper we present three problems involving some inequalities in three non-negative
real variables.

Problem 1. Let a, b, c be three non-negative real numbers such that ab + bc + ca > 0. Prove
that:

8a Å b − c ã 2 8b Å c − a ã 2 8c Å a − b ã 2
+ + + + + ≥ 6.
b + c b + c c + a c + a a + b a + b

Solution. We first prove the following three inequalities.

… …
a b c abc
+ + ≥ 2 1 + . (1)
b + c c + a a + b (a + b) (b + c) (c + a)
Indeed, using Minkowski’s Inequality and Schur’s Inequality we have

»
P p 3
2
a (a + b + c) + abc (a + b + c) + 9abc
, so
LHS = p ≥ p
(a + b) (b + c) (c + a) (a + b) (b + c) (c + a)

P
4abc a (a − b) (a − c)
LHS ≥ 4 + +
(a + b) (b + c) (c + a) (a + b) (b + c) (c + a)

4abc
≥ 4 + = RHS.
(a + b) (b + c) (c + a)

# 2
" ñ ô
b − c c − a a − b b − c c − a a − b
Å ã 2 Å ã 2 Å ã 2 Å ã 2 Å ã 2 Å ã 2
+ + ≥ 2 + + . (2)
b + c c + a a + b b + c c + a a + b
Indeed, let
b − c c − a a − b
x = , y = , z = ,
b + c c + a a + b
so
x + y + z + xyz = 0, −1 ≤ x, y, z ≤ 1.

We need to prove
Ä√ p √ ä 2
2
2
2
2
x + y + z 2 ≥ 2 x + y + z 2 i.e.
1
Teacher, VIET AU High School, District 12, Ho Chi Minh City, Vietnam, duyanh175@gmail.com
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