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

2.4 Establishing New Bounds
We first give two alternative ways to prove Nesbitt’s inequality.
(Nesbitt) For all positive real numbers a, b, c, we have
a
b + c +
(
2
a
Proof 4. From
b+c 2) − 1 ≥ 0, we deduce that
It follows that
Proof 5. We claim that
a
b + c ≥ 1 4 ·
∑
cyclic
b
c + a +
8a
b+c − 1
a
b+c + 1 =
a
b + c ≥ ∑
cyclic
c
a + b ≥ 3 2 .
8a − b − c
4(a + b + c) .
8a − b − c
4(a + b + c) = 3 2 .
a
b + c ≥ 3a 3 (
)
2
(
) or 2 a 3 3 3
2 + b 2 + c 2 ≥ 3a 1 2 (b + c).
2 a 3 2 + b 3 2 + c 3 2
The AM-GM inequality gives a 3 2 +b 3 2 +b 3 2 ≥ 3a 1 2 b and a 3 2 +c 3 2 +c 3 2 ≥ 3a 1 2
(
)
c . Adding these two inequalities
yields 2 a 3 2 + b 3 2 + c 3 2 ≥ 3a 1 2 (b + c), as desired. Therefore, we have
∑
cyclic
a
b + c ≥ 3 2
∑
cyclic
a 3 2
a 3 2 + b 3 2 + c 3 2
= 3 2 .
Some cyclic inequalities can be proved by finding new bounds. Suppose that we want to establish that
∑
F (x, y, z) ≥ C.
If a function G satisfies
cyclic
(1) F (x, y, z) ≥ G(x, y, z) for all x, y, z > 0, and
(2) ∑ cyclic
G(x, y, z) = C for all x, y, z > 0,
then, we deduce that
∑
F (x, y, z) ≥ ∑
G(x, y, z) = C.
cyclic
For example, if a function F satisfies
F (x, y, z) ≥
cyclic
x
x + y + z
for all x, y, z > 0, then, taking the cyclic sum yields
∑
F (x, y, z) ≥ 1.
cyclic
As we saw in the above two proofs of Nesbitt’s inequality, there are various lower bounds.
Problem 19. Let a, b, c be the lengths of a triangle. Show that
a
b + c +
b
c + a +
c
a + b < 2.
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