Advanced book on Mathematics Olympiad

140 3 Real Analysis

Example.Letf : R → Rbe a twice-differentiable function, with positive second
derivative. Prove that

f(x+f′(x))≥f(x),

for any real numberx.

Solution.Ifxis such thatf′(x)=0, then the relation holds with equality. If for a certain
x,f′(x) <0, then the mean value theorem applied on the interval[x+f′(x), x]yields

f(x)−f(x+f′(x))=f′(c)(−f′(x)),

for somecwithx+f′(x)<c<x. Because the second derivative is positive,f′is
increasing; hencef′(c)<f′(x) <0. Therefore,f(x)−f(x+f′(x)) <0, which
yields the required inequality.
In the casef′(x) >0, by the same argumentf(x+f′(x))−f(x)=f′(x)f′(c)forc
betweenxandx+f′(x), andf′(c)>f′(x) >0. We obtain againf(x)−f(x+f′(x)) <
0, as desired. 

Example.Find all real solutions to the equation

4 x+ 6 x

2

= 5 x+ 5 x

2

.

Solution.This problem was given at the 1984 Romanian Mathematical Olympiad, being
proposed by M. Chiri ̧t ̆a. The solution runs as follows.
Note thatx=0 andx=1 satisfy the equation from the statement. Are there other
solutions? The answer is no, but to prove it we use the amazing idea of treating the
numbers 4, 5 ,6 as variables and the presumably new solutionxas a constant.
Thus let us consider the functionf(t)=tx
2
+( 10 −t)x. The fact thatxsatisfies the
equation from the statement translates tof( 5 )=f( 6 ). By Rolle’s theorem there exists
c∈( 5 , 6 ), such thatf′(c)=0. This means thatx^2 cx

(^2) − 1
−x( 10 −c)x−^1 =0, or
xcx
(^2) − 1
=( 10 −c)x−^1.
Because exponentials are positive, this implies thatxis positive.
Ifx>1, thenxcx
(^2) − 1

cx
(^2) − 1
cx−^1 >( 10 −c)x−^1 , which is impossible since the
first and the last terms in this chain of inequalities are equal. Here we used the fact that
c>5.
If 0<x<1, thenxcx
(^2) − 1
<xcx−^1. Let us prove thatxcx−^1 <( 10 −c)x−^1. With the
substitutiony=x−1,y∈(− 1 , 0 ), the inequality can be rewritten asy+ 1 <(^10 c−c)y.
The exponential has base less than 1, so it is decreasing, while the linear function on the
left is increasing. The two meet aty=0. The inequality follows. Using it we conclude
again thatxcx
(^2) − 1
cannot be equal to( 10 −c)x−^1. This shows that a third solution to the
equation from the statement does not exist. So the only solutions to the given equation
arex=0 andx=1.