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

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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 ≤ yk + yk+2 for 1 ≤ k ≤ n − 2. Suppose further thatPn

k=1yk = 0. Prove that

n

X

k=1

k2yk ≥(n + 1)

n

X

k=1

kyk,

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+ yk)

=

n

X

k=1

k + 1 2

n − k 2



−2k 2

n − k + 1 2



+k − 1 2

n − k + 2 2



yk

=

n

X

k=1



3k2−3(n + 1)k +n + 2 2



yk

= 3

n

X

k=1

k2yk−(n + 1)

n

X

k=1

kyk

!

where in the last step we used the fact thatPn

k=1yk= 0.

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

Moreover, since k+12  n−k

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

By imposingPn

k=1yk = 0, we find that yk= 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).

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