Advanced book on Mathematics Olympiad

2.1 Identities and Inequalities 43

Example.Letα 1 ,α 2 ,...,αnbe positive real numbers,n≥2, such thatα 1 +α 2 +···+

αn=1. Prove that

α 1

1 +α 2 +···+αn

+

α 2

1 +α 1 +···+αn

+···+

αn

1 +α 1 +···+αn− 1

≥

n

2 n− 1

.

Solution.Rewrite the inequality as

α 1

2 −α 1

+

α 2

2 −α 2

+···+

αn

2 −αn

≥

n

2 n− 1

,

and then define the function

f(α 1 ,α 2 ,...,αn)=

α 1

2 −α 1

+

α 2

2 −α 2

+···+

αn

2 −αn

.

As said in the statement, this function is defined on the subset ofRnconsisting of points
whose coordinates are positive and add up to 1. We would like to show that on this set
fis greater than or equal to 2 nn− 1.
Doesfhave a minimum? The domain offis bounded but isnotclosed, being the
interior of a tetrahedron. We can enlarge it, though, by adding the boundary, to the set

M={(α 1 ,α 2 ,...,αn)|α 1 +α 2 +···+αn= 1 ,αi≥ 0 ,i= 1 , 2 ,...,n}.

We now know thatfhas a minimum onM.

A look at the original inequality suggests that the minimum is attained when all the

αi’s are equal. So let us choose a point(α 1 ,α 2 ,...,αn)for whichαi   =αjfor some

indicesi, j. Assume thatαi<αjand let us see what happens if we substituteαi+xfor

αiandαj−xforαj, with 0<x<αj−αi. In the defining expression off, only the

ith andjth terms change. Moreover,

αi

2 −αi

+

αj

2 −αj

−

αi+x

2 −αi−x

−

αj−x

2 −αj+x

=

2 x(αj−αi−x)( 4 −αi−αj)

( 2 −αi)( 2 −αj)( 2 −αi−x)( 2 −αj−x)

> 0 ,

so on moving the numbers closer, the value offdecreases. It follows that the point that

we picked was not a minimum. Hence the only possible minimum is(^1 n,^1 n,...,n^1 ),in

which case the value offis 2 nn− 1. This proves the inequality. 

However, in most situations, as is the case with this problem, we can bypass the use

of real analysis and argue as follows. If theai’s were not all equal, then one of them must

be less than^1 nand one of them must be greater. Take these two numbers and move them

closer until one of them reaches^1 n. Then stop and choose another pair. Continue the

algorithm until all numbers become^1 n. At this very moment, the value of the expression