Inequality aq + br + cp < k^2

[Guest]

Suppose that [tex]a, b, c, p, q, r > 0[/tex] and [tex]a+ p = b + q = c + r = k[/tex].
Prove that
[tex]aq + br + cp < k^2[/tex].

[Guest]

k^3 = (a+p))b+q)(c+q)

k^3= abc + abr + aqc + aqr+ bpc + bpr + pqc + pqr

k^3= aq(r+c) + br (a+p) + (cp)(b+q) + abc + pqr
k^3= k(aq + br+ cp) + abc + pqr
k^3> k(aq+br+cp) [abc, pqr>0]
k^2 > aq+br+cp
Enjoy

[Guest]

Thanks!

[Guest]

no problem. I know its hella late tho lmaoooo