Advanced book on Mathematics Olympiad

10 1 Methods of Proof

The inequality is now obvious sincea 1 a 2 ≥1 anda 1 +a 2 ≥ 2

√

a 1 a 2.
Now instead of exhausting all positive integersn, we downgrade our goal and check
just the powers of 2. So we prove that the inequality holds forn= 2 kby induction onk.
Assuming it true fork, we can write

(^2) ∑k+^1
i= 1

1

1 +ai

=

∑^2 k

i= 1

1

1 +ai

+

(^2) ∑k+^1
i= 2 k+ 1

1

1 +ai

≥ 2 k

(

1

1 +^2 k

√

a 1 a 2 ···a 2 k

+

1

1 +^2 k

√

a 2 k+ 1 a 2 k+ 2 ···a 2 k+ 1

)

≥ 2 k

2

1 +^2 k+^1

√

a 1 a 2 ···a 2 k+ 1

,

where the first inequality follows from the induction hypothesis, and the second is just

the base case. This completes the induction.

Now we have to cover the cases in whichnis not a power of 2. We do the induction

backward, namely, we assume that the inequality holds forn+1 numbers and prove it

forn. Leta 1 ,a 2 ,...,anbe some real numbers greater than 1. Attach to them the number

√na 1 a 2 ···an. When writing the inequality for thesen+1 numbers, we obtain

1

1 +a 1

+···+

1

1 +n

√

a 1 a 2 ···an

≥

n+ 1

1 +n+^1

√

a 1 ···ann

√

a 1 a 2 ···an

.

Recognize the complicated radical on the right to be n

√

a 1 a 2 ···an. After cancelling the

last term on the left, we obtain

1

1 +a 1

+

1

1 +a 2

+···+

1

1 +an

≥

n

1 +n

√

a 1 a 2 ···an

,

as desired. The inequality is now proved, since we can reach any positive integernby

starting with a sufficiently large power of 2 and working backward. 

Try to apply the same technique to the following problems.

31.Letf :R→Rbe a function satisfyingf(x^1 + 2 x^2 )=f(x^1 )+ 2 f(x^2 )for anyx 1 ,x 2.

Prove that

f

(

x 1 +x 2 +···+xn

n

)

=

f(x 1 )+f(x 2 )+···+f(xn)

n

for anyx 1 ,x 2 ,...,xn.

32.Show that ifa 1 ,a 2 ,...,anare nonnegative numbers, then

( 1 +a 1 )( 1 +a 2 )···( 1 +an)≥( 1 +n

√

a 1 a 2 ···an)n.