In a quadrilateral PQRS, the diagonals PR and QS intersect each other at O. Show that
(i) PQ + QR + RS + SP > PR + QS.
(ii) PQ + QR + RS + SP < 2 (PR + QS). 

We know that in a triangle the sum of two sides of a triangle is greater than the third side. Therefore in ΔPSR, we have   
    
PS + SR > PR   ...... (i)
In ΔQSR, QR + RS > QS  ...... (ii)
In ΔPQS, PQ + SP > QS   ...... (iii)
In ΔPQR, PQ + QR > PR    ..... (iv)
Adding (i), (ii), (iii) and (iv), we get
2 (SP + PQ + QR + RS) > 2 (QS + PR)  PQ + QR + RS + SP > PR + QS
Similarly in ΔPQO, ΔQRO, ΔRSO and ΔSPO,  
we have
                 QO + PO > PQ ...... (v)
                 QO + RO > QR ...... (vi)
                 RO + SO > RS ...... (vii)
                 SO + PO > SP ...... (viii)
Adding (v), (vi), (vii) and (viii),  
we get
2 (QO + SO + PO + OR) > PQ + QR + RS +SP 2 (QS + PR) > PQ + QR + RS + SP
  or PQ + QR + RS + SP < 2 (QS + PR).