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Z 1 0 fa(x)gb(x) dx  Z 1 0 fb(x)ga(x) dx

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Problem 11819

(American Mathematical Monthly, Vol.122, February 2015) Proposed by C. Lupu (USA).

Let f be a continuous, nonnegative function on [0, 1]. Show that Z 1

0

f3(x) dx ≥ 4

Z 1 0

x2f (x) dx

 Z 1 0

xf2(x) dx

 .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We will show a more general result.

Let f and g be continuous, nonnegative functions on [0, 1] and let a and b be nonnegative real numbers. Then

Z 1 0

fa+b(x) dx

 Z 1 0

ga+b(x) dx



Z 1 0

fa(x)gb(x) dx

 Z 1 0

fb(x)ga(x) dx

 .

The required inequality follows by taking g(x) = x, a = 2, and b = 1.

Let A, B be nonnegative real numbers then

(Aa− Ba)(Ab− Bb) ≥ 0 which implies that

Aa+b+ Ba+b≥ AaBb+ AbBa.

Let A = f (x)g(y) and B = f (y)g(x) then by integrating over [0, 1] × [0, 1] we obtain Z 1

x=0

Z 1 y=0

(f (x)g(y))a+bdx dy + Z 1

x=0

Z 1 y=0

(f (y)g(x))a+bdx dy

≥ Z 1

x=0

Z 1 y=0

(f (x)g(y))a(f (y)g(x))bdx dy + Z 1

x=0

Z 1 y=0

that is

Z 1 0

fa+b(x) dx

 Z 1 0

ga+b(y) dy

 +

Z 1 0

fa+b(y) dy

 Z 1 0

ga+b(x) dx



Z 1 0

fa(x)gb(x) dx

 Z 1 0

fb(y)ga(y) dy

 +

Z 1 0

fa(y)gb(y) dy

 Z 1 0

fb(x)ga(x) dx

 ,

and the proof is complete. 

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