If px2 + qx + r = 0 has no real roots and p, q, r are real such that p + r > 0 then
p - q + r > 0
Let α+iβ,α−iβ be the roots. Then α2+β2=rp>0. So p, r are of the same sign. Also p + r > 0. So p, r are both positive.
If q < 0, p - q + r > 0
If q > 0, (p+r)2−(p−r)2=4pr≥q2(∵ roots are nonreal)
∴(p+r)2≥q2+(p−r)2≥
q2 ∴ p + r > q