0 then n X k=1 k Pk j=11/aj ≤ c n X k=1 ak

Problem 11145

(American Mathematical Monthly, Vol.112, April 2005) Proposed by J. Zinn (USA).

Find the leastc such that if n ≥ 1 and a¹, . . . , an > 0 then

n

X

k=1

k Pk

j=11/aj

≤ c

n

X

k=1

ak.

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

Let hk = k/Pk

j=11/aj i. e. the harmonic mean of a¹, . . . , ak. If we take ak= 1/k we have that

n

X

k=1

hk =

n

X

k=1

k 1 + · · · + k =

n

X

k=1

2 k + 1 = 2



Hn− n n + 1



where Hn =Pn

k=11/k is the n-th harmonic number. Hence

c ≥

n

X

k=1

hk

! /

n

X

k=1

ak

!

= 2



1 − n

(n + 1)Hn

 ,

and, since Hn goes to infinity, the above inequality holds for every positive integer n only if the constant c is greater or equal to 2.

Now we prove that it actually holds for c = 2 and therefore this is just the best constant. Since h²_k+ h²_k−1≥ 2hkh_k−1

then

hk− h²_k 2ak

= hk−h²_k 2

 k

hk −k − 1 h_k−1



= hk

 1 −k

2



+k − 1 2

h²_k h_k−1

≥ hk

 1 − k

2



+k − 1

2 (2hk− hk−1) = 1

2(khk− (k − 1)hk−1) . Summing up for k = 1, . . . , n (let h⁰= 0) we find that

n

X

k=1

hk−1 2

n

X

k=1

h²_k ak ≥ 1

2

n

X

k=1

(khk− (k − 1)hk−1) = 1

2nhn≥ 0 hence

2

n

X

k=1

hk ≥

n

X

k=1

 hk

√ak

² .

After multipling byPn

k=1ak, by Cauchy’s inequality we obtain

2

n

X

k=1

ak·

n

X

k=1

hk≥

n

X

k=1

(√ ak)²·

n

X

k=1

 hk

√ak

²

≥

n

X

k=1

hk

!²

that is

2

n

X

k=1

ak ≥

n

X

k=1

hk.

It is interesting to remark that the proposed inequality is a variation (take p = −1) of a result due to G. H. Hardy (see for example the book Inequalities of Hardy-Littlewood-Polya): for p > 1

n

X

k=1



 1 k

k

X

j=1

a¹_j^/p





p

≤

 p p − 1

p n

X

k=1

ak.