5.2 ALGEBRAIC MANIPULATION REVISITED 155These tenns look suspiciously like they came from squares of binomials. For
example,
(XIX 2 -XIX 4 )^2 = xTx� -2xTx 2 X 4 +xT�
produces two of the perfect square tenns and one cross-tenn. Our strategy: Write the
sum of left-hand sides in the fonn
(^1) ( 2 2 2 )
"2 YI +Y 2 +···+YIO ,
where each Yk produces a different cross-tenn, and all the perfect square terms are
exactly duplicated. And indeed, after some experimentation, we have
s
o � }:. (x^7 -Xi+ 2 Xi+4)(x^7 + 1 -Xi+ 2 Xi+ 4 )
1 = 1
1 � ( 2 2 )
= "24. (XiXi+ 1 -XiXi+ 3 ) + (Xi-IXi+ 1 -Xi-IXi+ 3 ).
1= 1
Since we have written 0 as greater than or equal to a sum of squares, the only solution
is when all the squares are zero, and this implies that
XI =X 2 =X 3 =X 4 =Xs·
Consequently, the solution set to the system of inequalities is
{(u, u,u ,u,u) : u E lR}. •
The final example looks at a tricky inequality. We don't solve it, but a nice substi
tution makes it at least somewhat more tractable.
Example 5.2.23 (IMO 19 95) Let a,b,e be positive real numbers such that abc = 1.
Prove that
Partial Solution: We will not solve this problem just yet, but point out an alge
braic simplification that must be done. What is the worst thing about this problem? It
is an inequality involving fairly ugly fractions. Wishful thinking tells us that it would
be nicer if the fractions either were less ugly or did not exist at all. How can this be
achieved? There is a pretty obvious substitution -but only obvious if you have the
idea of substitution in the forefront of your consciousness. The substitution is
X = 1la,y = 1/b,z = lie,
which transfonns the original inequality (use the fact that xyz = 1) into
x^2 y^2 z^2 3
- +-+-->-.
y+z z+x x+y - 2
This inequality is still not that easy to deal with, but the denominators are much less
complicated, and the problem has been reduced in complexity. See Example 5.5. 23
for continuation.