Advanced book on Mathematics Olympiad

516 Real Analysis

f(x)=(x+y+z)ln(x+y+z)+xlnx+ylny+zlnz

−(x+y)ln(x+y)−(y+z)ln(y+z)−(z+x)ln(z+x).

Differentiatingf(x)with respect tox, we obtain

f′(x)=ln

(x+y+z)x

(x+y)(z+x)

=ln

x^2 +yx+zx

x^2 +yx+zx+yz

<ln 1= 0 ,

for all positive numbersx. It follows thatf(x)is strictly decreasing, sof(x) <

limt→ 0 f(t)=0, for allx>0. Henceef(x)<1 for allx>0, which is equivalent to the

first inequality from the statement.

(b) We apply the same idea, fixingy, z >0 and considering the functiong:( 0 ,∞)

→R,

g(x)=(x+y+z)^2 ln(x+y+z)+x^2 lnx+y^2 lny+z^2 lnz

−(x+y)^2 ln(x+y)−(y+z)^2 ln(y+z)−(z+x)^2 ln(z+x).

Differentiating with respect tox, we obtain

g′(x)=2ln

(x+y+z)x+y+zxx

(x+y)x+y(z+x)z+x

.

We would like to show this time thatgis increasing, for theng(x) >limt→ 0 g(t)=0,

from which the desired inequality is obtained by exponentiation. We are left to prove

thatg′(x) >0, which is equivalent to

(x+y+z)x+y+zxx>(x+y)x+y(z+x)z+x, forx, y, z > 0.

And we take the same path as in (a). Because we want to make the derivative as simple
as possible, we fixx, y >0 and defineh:( 0 ,∞)→R,

h(z)=(x+y+z)ln(x+y+z)+xlnx−(x+y)ln(x+y)−(z+x)ln(z+x).

Then

h′(z)=ln

x+y+z

z+x

>ln 1= 0 ,

forz>0. Henceh(z) >limt→ 0 h(t)=0,z>0. This implies the desired inequality

and completes the solution.

(American Mathematical Monthly, proposed by Sz. András, solution by H.-J. Seiffert)

413.Let us examine the functionF(x)=f(x)−g(x). BecauseF(n)(a)   =0, we have

F(n)(x) =0 forxin a neighborhood ofa. HenceF(n−^1 )(x)  =0 forx =aandxin a

neighborhood ofa(otherwise, this would contradict Rolle’s theorem). ThenF(n−^2 )(x)