sorry
I suspect you have already tried expanding all that out
From
(p-q)^2+(q-r)^2+(r-p)^2>(pq+qr+rp)
hence deduce that for any number
p,q,r,(p+q+r)>3(pq+qr+rp)
4 answers
Sir am too confuse at what am doing do u have any ideal at all that might help?
(p-q)^2 = p^2 - 2 pq + q^2
(q-r)^2 = q^2 - 2 qr + r^2
(r-p)^2 = r^2 - 2 rp + p^2
---------------------------add them
2 p^2 + 2 r^2+ 2 q^2 - 2(pq+qr+rp)
= 2[ p^2 + r^2+ q^2 -(pq+qr+rp)]
well all those squared quantities are positive so
that quantity is > (pq+qr+rp)]
(q-r)^2 = q^2 - 2 qr + r^2
(r-p)^2 = r^2 - 2 rp + p^2
---------------------------add them
2 p^2 + 2 r^2+ 2 q^2 - 2(pq+qr+rp)
= 2[ p^2 + r^2+ q^2 -(pq+qr+rp)]
well all those squared quantities are positive so
that quantity is > (pq+qr+rp)]
God bless your damon and may your knowledge be increased in jesus name
1 answer