Advanced book on Mathematics Olympiad

40 2 Algebra

Setting this equal to zero, we find the unique critical pointx= n−^1

√

x 1 x 2 ···xn, since in

this casex^1 −

(^1) n
= n

√

x 1 x 2 ···xn− 1. Moreover, the functionx^1 −

(^1) n
is increasing on( 0 ,∞);
hencef′(x) <0 forx<n−^1

√

x 1 x 2 ···xn− 1 , andf′(x) >0 forx>n−^1

√

x 1 x 2 ···xn− 1 .We
find thatfhas a global minimum atx=n−^1

√

x 1 x 2 ···xn− 1 , where it takes the value

f(n−^1

√

x 1 x 2 ···xn− 1 )=

x 1 +x 2 +···+xn− 1 +n−^1

√

x 1 x 2 ···xn− 1

n

−n

√

x 1 x 2 ···xn− 1 ·n(n−^1 )

√

x 1 x 2 ···xn− 1

=

x 1 +x 2 +···+xn− 1 +n−^1

√

x 1 x 2 ···xn− 1

n

−n−^1

√

x 1 x 2 ···xn− 1

=

x 1 +x 2 +···+xn− 1 −(n− 1 )n−^1

√

x 1 x 2 ···xn− 1

n

.

By the induction hypothesis, this minimum is nonnegative, and is equal to 0 if and only
ifx 1 =x 2 = ··· =xn− 1. We conclude thatf(xn)≥0 with equality if and only if
x 1 =x 2 = ··· =xn− 1 andxn=n−√^1 x 1 x 2 ···xn− 1 =x 1. This completes the induction.
We apply the AM–GM inequality to solve two problems composed by the second
author of the book.

Example.Find the global minimum of the functionf:R^2 →R,

f (x, y)= 3 x+y( 3 x−^1 + 3 y−^1 − 1 ).

Solution.The expression

3 f (x, y)+ 1 = 32 x+y+ 3 x+^2 y+ 1 − 3 · 3 x+y

is of the forma^3 +b^3 +c^3 − 3 abc, wherea=^3

√

32 x+y,b=^3

√

3 x+^2 y, andc=1, all
of which are positive. By the AM–GM inequality, this expression is nonnegative. It is
equal to zero only whena=b=c, that is, when 2x+y=x+ 2 y=0. We conclude
that the minimum offisf( 0 , 0 )=−^13. 

Example.Leta, b, c, dbe positive real numbers withabcd=1. Prove that

a

b+c+d+ 1

+

b

c+d+a+ 1

+

c

d+a+b+ 1

+

d

a+b+c+ 1

≥ 1.

Solution.A first idea is to homogenize this inequality, and for that we replace the 1 in
each denominator by^4

√

abcd, transforming the inequality into

a

b+c+d+^4

√

abcd

+

b

c+d+a+^4

√

abcd

+

c

d+a+b+^4

√

abcd

+

d

a+b+c+^4

√

abcd

≥ 1.