Advanced book on Mathematics Olympiad

(ff) #1
Real Analysis 515

−cos^2 xcos^2 (sinx)cos(sin(sin(sinx)))cos(sin(sin(sin(sinx))))sin(sin(sinx))
−cos^2 xcos^2 (sinx)cos^2 (sin(sinx))cos(sin(sin(sin(sinx))))sin(sin(sin(sinx)))
−cos^2 xcos^2 (sinx))cos^2 (sin(sinx))cos^2 (sin(sin(sinx)))sin(sin(sin(sin(sinx)))),

which is clearly nonpositive for 0 ≤ x ≤ 1. This means thatf′(x)is monotonic.
Therefore,f′(x)has at most one rootx′in[ 0 ,+∞).Thenf(x)is monotonic at[ 0 ,x′]
and[x′,+∞)and has at most two nonnegative roots. Becausef(x)is an odd function,
it also has at most two nonpositive roots. Therefore,−x 0 , 0 ,x 0 are the only solutions.


410.Define the functionG:R→R,G(x)=(


∫x
0 f(t)dt)

(^2). It satisfies
G′(x)= 2 f(x)
∫x
0
f(t)dt.
BecauseG′( 0 )=0 andG′(x)=g(x)is nonincreasing it follows thatG′is nonnegative on
(−∞, 0 )and nonpositive on( 0 ,∞). This implies thatGis nondecreasing on(−∞, 0 )
and nonincreasing on( 0 ,∞). And this, combined with the fact thatG( 0 )=0 and
G(x)≥0 for allx, impliesG(x)=0 for allx. Hence
∫x
0 f(t)dt=0. Differentiating
with respect tox, we conclude thatf(x)=0 for allx, and we are done.
(Romanian Olympiad, 1978, proposed by S. R ̆adulescu)
411.Consider the function
F(t)=
[∫t
0
f(x)dx


] 2


∫t

0

[f(x)]^3 dx fort∈[ 0 , 1 ].

We want to show thatF(t)≥0, from which the conclusion would then follow. Because
F( 0 )=0, it suffices to show thatFis increasing. To prove this fact we differentiate and
obtain


F′(t)=f(t)

[

2

∫t

0

f(x)dx−f^2 (t)

]

.

It remains to check thatG(t)= 2


∫t
0 f(x)dx−f

(^2) (t)is positive on[ 0 , 1 ]. Because
G( 0 )=0, it suffices to prove thatGitself is increasing on[ 0 , 1 ]. We have
G′(t)= 2 f(t)− 2 f(t)f′(t).
This function is positive, since on the one handf′( 0 )≤1, and on the other handf
is increasing, having a positive derivative, and sof(t)≥f( 0 )=0. This proves the
inequality. An example in which equality holds is the functionf:[ 0 , 1 ]→R,f(x)=x.
(34th W.L. Putnam Mathematical Competition, 1973)
412.(a) To avoid the complicated exponents, divide the inequality by the right-hand side;
then take the natural logarithm. Next, fix positive numbersyandz, and then introduce
the functionf:( 0 ,∞)→R,

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