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

Proof. Since x, y, z ≥ 0, we can find positive sequences {x n }, {y n }, {z n } for which
Applying the theorem 2 yields
lim x n = x,
n→∞
lim y n = y, lim z n = z.
n→∞ n→∞
x n y n z n ≥ (y n + z n − x n )(z n + x n − y n )(x n + y n − z n ).
Now, taking the limits to both sides, we get the result.
Clearly, the equality holds when x = y = z. However, xyz = (y+z −x)(z +x−y)(x+y−z) and x, y, z ≥ 0
does not guarantee that x = y = z. In fact, for x, y, z ≥ 0, the equality xyz = (y +z −x)(z +x−y)(x+y −z)
is equivalent to
x = y = z or x = y, z = 0 or y = z, x = 0 or z = x, y = 0.
It’s straightforward to verify the equality
xyz − (y + z − x)(z + x − y)(x + y − z) = x(x − y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y).
Hence, the theorem 4 is a particular case of Schur’s inequality.
Problem 1. (IMO 2000/2, Proposed by Titu Andreescu) Let a, b, c be positive numbers such that
abc = 1. Prove that (
a − 1 + 1 ) (
b − 1 + 1 ) (
c − 1 + 1 )
≤ 1.
b
c
a
First Solution. Since abc = 1, we make the substitution a = x y , b = y z , c = z x
for x, y, z > 0.3 We rewrite
the given inequality in the terms of x, y, z :
( x
y − 1 + z ) (y
y z − 1 + x ) ( z
z x x)
− 1 + y ≤ 1 ⇔ xyz ≥ (y + z − x)(z + x − y)(x + y − z).
The Ravi Substitution is useful for inequalities for the lengths a, b, c of a triangle. After the Ravi
Substitution, we can remove the condition that they are the lengths of the sides of a triangle.
Problem 2. (IMO 1983/6) Let a, b, c be the lengths of the sides of a triangle. Prove that
a 2 b(a − b) + b 2 c(b − c) + c 2 a(c − a) ≥ 0.
First Solution. After setting a = y + z, b = z + x, c = x + y for x, y, z > 0, it becomes
x 3 z + y 3 x + z 3 y ≥ x 2 yz + xy 2 z + xyz 2
or x2
y + y2
z + z2
x ≥ x + y + z,
which follows from the Cauchy-Schwarz inequality
( )
x
2
(y + z + x)
y + y2
z + z2
≥ (x + y + z) 2 .
x
Exercise 2. Let a, b, c be the lengths of a triangle. Show that
a
b + c +
b
c + a +
c
a + b < 2.
3 For example, take x = 1, y = 1 a , z = 1 ab . 2