Topics in Inequalities - Theorems and Techniques Hojoo ... - Index of

Problem 14. Let a, b, c be positive real numbers satisfying a + b + c = 1. Show that
Solution. We want to establish that
Set x =
√
bc
a , y = √ ca
b , z = √
a
a + bc + b
√
abc
b + ca + c + ab ≤ 1 + 3√ 3
4 .
1
1 + bc a
+ 1
1 + ca b
+
√
ab
c
1 + ab
c
ab
c
. We need to prove that
≤ 1 + 3√ 3
4 .
1
1 + x 2 + 1
1 + y 2 + z
1 + z 2 ≤ 1 + 3√ 3
4 ,
where x, y, z > 0 and xy + yz + zx = 1. It’s not hard to show that there exists A, B, C ∈ (0, π) with
The inequality becomes
x = tan A 2 , y = tan B 2 , z = tan C , and A + B + C = π.
2
or
or
1
1 + ( 1
)
tan A 2
+
2
1 + ( tan C 2
)
tan B 2
+
2
1 + ( tan C 2
1 + 1 2 (cos A + cos B + sin C) ≤ 1 + 3√ 3
4
cos A + cos B + sin C ≤ 3√ 3
2 .
) 2
≤ 1 + 3√ 3
4
Note that cos A + cos B = 2 cos ( ) (
A+B
2 cos A−B
) ∣ 2 . Since A−B ∣
2
< π 2
, this means that
( ) ( )
A + B
π − C
cos A + cos B ≤ 2 cos = 2 cos .
2
2
It will be enough to show that
( ) π − C
2 cos + sin C ≤ 3√ 3
2
2 ,
where C ∈ (0, π). This is a one-variable inequality. 3 It’s left as an exercise for the reader.
Problem 15. (Iran 1998) Prove that, for all x, y, z > 1 such that 1 x + 1 y + 1 z = 2,
√ x + y + z ≥
√
x − 1 +
√
y − 1 +
√
z − 1.
First Solution. We begin with the algebraic substitution a = √ x − 1, b = √ y − 1, c = √ z − 1. Then, the
condition becomes
1
1 + a 2 + 1
1 + b 2 + 1
1 + c 2 = 2 ⇔ a2 b 2 + b 2 c 2 + c 2 a 2 + 2a 2 b 2 c 2 = 1
and the inequality is equivalent to
√
a2 + b 2 + c 2 + 3 ≥ a + b + c ⇔ ab + bc + ca ≤ 3 2 .
Let p = bc, q = ca, r = ab. Our job is to prove that p + q + r ≤ 3 2 where p2 + q 2 + r 2 + 2pqr = 1. By the
exercise 7, we can make the trigonometric substitution
(
p = cos A, q = cos B, r = cos C for some A, B, C ∈ 0, π )
with A + B + C = π.
2
What we need to show is now that cos A + cos B + cos C ≤ 3 2
. It follows from Jensen’s inequality.
3
Differentiate! Shiing-shen Chern
18