n, then, by Cauchy-Schwarz inequality, |zj||zk| =q |xj|2+ |yj|2·p|xk|2+ |yk|2≥ |xj||xk

Problem 11840

(American Mathematical Monthly, Vol.122, May 2015) Proposed by G. Stoica (Canada).

Let z1, . . . , zn be complex numbers. Prove that

n

X

k=1

|z_k|

!2

−     

n

X

k=1

z_k     

2

≥

n

X

k=1

|Re(z_k)| −     

n

X

k=1

Re(z_k)     

!2

.

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

Let z_k= x_k+ iy_k for k = 1, . . . , n, then, by Cauchy-Schwarz inequality,

|zj||zk| =q

|xj|²+ |yj|²·p|xk|²+ |yk|²≥ |xj||xk| + |yj||yk| ≥ |xj||xk| + yjyk. Hence

n

X

k=1

|zk|

!²

−     

n

X

k=1

zk

2

= 2 X

1≤j<k≤n

|zj||zk| − 2 X

1≤j<k≤n

Re(zjzk)

= 2 X

1≤j<k≤n

(|zj||zk| − (xjxk+ yjyk))

≥ 2 X

1≤j<k≤n

(|xj||xk| − xjxk)

=

n

X

k=1

|xk|

!²

−     

n

X

k=1

xk

2

=

n

X

k=1

|x_k| +     

n

X

k=1

x_k     

! _n X

k=1

|x_k| −     

n

X

k=1

x_k     

!

≥

n

X

k=1

|xk| −     

n

X

k=1

xk

!²

=

n

X

k=1

|Re(zk)| −     

n

X

k=1

Re(zk)     

!² ,

where we also applied the inequalityPn

k=1|xk| ≥ |Pn

k=1xk|.