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)
sorry
I suspect you have already tried expanding all that out
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)]
God bless your damon and may your knowledge be increased in jesus name
To prove the inequality for any numbers p, q, and r, we will start with the given expression:
(p-q)^2 + (q-r)^2 + (r-p)^2 > (pq + qr + rp)
Expanding the squares:
p^2 - 2pq + q^2 + q^2 - 2qr + r^2 + r^2 - 2rp + p^2 > pq + qr + rp
Combining like terms:
2p^2 + 2q^2 + 2r^2 - 2pq - 2qr - 2rp > pq + qr + rp
Now, let's simplify this inequality further:
2(p^2 + q^2 + r^2 - pq - qr - rp) > (pq + qr + rp)
Dividing both sides by 2:
p^2 + q^2 + r^2 - pq - qr - rp > (pq + qr + rp) / 2
Re-arranging terms:
p^2 + q^2 + r^2 - pq - qr - rp - (pq + qr + rp) / 2 > 0
Combining the terms with a common denominator:
(2p^2 + 2q^2 + 2r^2 - 2pq - 2qr - 2rp - pq - qr - rp) / 2 > 0
Simplifying the numerator:
(2p^2 + p^2 + 2q^2 + q^2 + 2r^2 + r^2 - 2pq - pq - 2qr - qr - 2rp - rp) / 2 > 0
(3p^2 - 3pq + 3q^2 - 3qr + 3r^2 - 3rp) / 2 > 0
Multiplying both sides by 2:
3p^2 - 3pq + 3q^2 - 3qr + 3r^2 - 3rp > 0
Factoring out a 3:
3(p^2 - pq + q^2 - qr + r^2 - rp) > 0
Now, for the inequality to hold true, we know that the expression within the parentheses must be greater than 0:
p^2 - pq + q^2 - qr + r^2 - rp > 0
And if we simplify this further:
p(p - q) + q(q - r) + r(r - p) > 0
Now, since this inequality holds true for any values of p, q, and r, we can conclude that:
(p + q + r)(p - q + q - r + r - p) > 0
Simplifying:
(p + q + r)(0) > 0
Therefore, the resulting inequality is:
(p + q + r) > 0
Now, to deduce that (p + q + r) > 3(pq + qr + rp), we need to introduce another inequality:
3(p^2 - pq + q^2 - qr + r^2 - rp) > 0
Expanding the square:
3p^2 - 3pq + 3q^2 - 3qr + 3r^2 - 3rp > 0
Factoring out a common 3:
3(p^2 + q^2 + r^2 - pq - qr - rp) > 0
And since we proved earlier that (p^2 + q^2 + r^2 - pq - qr - rp) > 0, we can conclude that:
3(p^2 + q^2 + r^2 - pq - qr - rp) > 3(0)
3(p^2 + q^2 + r^2 - pq - qr - rp) > 0
This implies that:
3(p + q + r)(p - q + q - r + r - p) > 0
(p + q + r) > 0
Therefore, we can deduce that for any numbers p, q, r, (p + q + r) > 3(pq + qr + rp).