We have that n−2 X k=1 k + 1 2 n − k 2  (yk+2−2yk+1+ yk

Problem 11886

(American Mathematical Monthly, Vol.123, January 2016) Proposed by F. Holland (Ireland).

Suppose n ≥ 3, and let y1, . . . , yⁿ be a list of real numbers such that 2yk+1 ≤ y^k + yk+2 for 1 ≤ k ≤ n − 2. Suppose further thatPⁿ

k=1y^k = 0. Prove that

n

X

k=1

k²y^k ≥(n + 1)

n

X

k=1

ky^k,

and determine when equality holds.

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

We have that

n−2

X

k=1

k + 1 2

n − k 2



(yk+2−2yk+1+ y^k)

=

n

X

k=1

k + 1 2

n − k 2



−2k 2

n − k + 1 2



+k − 1 2

n − k + 2 2



y^k

=

n

X

k=1



3k²−3(n + 1)k +n + 2 2



y^k

= 3

n

X

k=1

k²y^k−(n + 1)

n

X

k=1

ky^k

!

where in the last step we used the fact thatPⁿ

k=1y^k= 0.

Hence, the inequality follows from the hypothesis yk+2−2yk+1+ y^k ≥0.

Moreover, since ^k+1_2
n−k

2
> 0, the equality holds if and only if yk+2−2yk+1+ y^k = 0 for 1 ≤ k ≤ n − 2, that is yk+2−yk+1= yk+1−y^k which implies that y^k= m(k − 1) + q with m, q ∈ R.

By imposingPⁿ

k=1y^k = 0, we find that y^k= m(k −ⁿ⁺¹₂ ) for 1 ≤ k ≤ n.  Remark. The following statement gives the continuous analogue of the above inequality.

Let f ∈ C²([0, 1]) be a real-valued convex function such thatR1

0 f (x) dx = 0 then Z ¹

0

x²f (x) dx ≥ Z ¹

0

xf (x) dx and the equality holds if and only if f (x) = m(x − 1/2) with m ∈ R.

Infact, by using integration by parts twice, we have that Z ¹

0

x²(x − 1)²

4 f^′′(x) dx = − Z ¹

0

x(x − 1)(2x − 1)

2 f^′(x) dx

= Z ¹

0



3x²−3x +1 2



f (x) dx = 3

Z 1 0

x²f (x) dx − Z ¹

0

xf (x) dx

 .

Hence, the inequality follows from the convexity hypothesis f^′′(x) ≥ 0. Moreover, since x²(x−1)²> 0 in (0, 1), the equality holds if and only if f^′′(x) = 0 in (0, 1) and, together with R¹

0 f (x) dx = 0, we obtain that f (x) = m(x − 1/2).