Since a2n+1= 0 and a2&gt

Problem 11861

(American Mathematical Monthly, Vol.122, October 2015) Proposed by Phu Cuong Le Van (Vietnam).

Letn be a natural number and let f be a continuous function from [0, 1] to R such thatR1

0 f (x)²ⁿ⁺¹dx = 0. Prove that

(2n + 1)²ⁿ⁺¹ (2n)²ⁿ

Z 1 0

f (x)dx

⁴ⁿ

≤ Z 1

0

(f (x))⁴ⁿdx.

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

For t ∈ R, by Cauchy-Schwarz inequality,

Z 1 0

(t + f (x)²ⁿ)f (x)dx

²

≤ Z 1

0

(t + f (x)²ⁿ)²dx Z 1

0

f (x)²dx,

Let a^k =R1

0 f (x)^kdx and assume that f is not identically zero (otherwise the inequality is trivial).

Since a2n+1= 0 and a2> 0, it follows that

a²₁t²= (a1t + a2n+1)²≤(t²+ 2a2nt + a4n)a2, that is

 a²₁ a2

−1



t²−2a2nt ≤ a4n.

By Cauchy-Schwarz inequality a²1≤a2. More precisely, we have that a²1< a2because equality holds iff f is constant which is in contradiction with the facts: a2n+1= 0 and a2> 0.

Now the LHS is a concave quadratic function with respect to t and it attains its maximum value at t = a2n/_a2

a1 2 −1

. Hence, we get

a²ⁿ⁺¹2

(a2−a²₁)

CS

≤ a²2n

1 − ^aa²¹

2

 ≤ a4n.

The LHS is a convex function with respect to a2 for a2 > a²₁ and it attains its minimum value at a2= (2n + 1)a²₁/(2n). Therefore

(2n + 1)²ⁿ⁺¹

(2n)²ⁿ a⁴ⁿ1 = ((2n + 1)a²1/(2n))²ⁿ⁺¹ ((2n + 1)a²₁/(2n) − a²₁) ≤a4n

and the proof is complete.