Problem
ALG-B2-M06-P017 Square of q
#17
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[(ab+bc+ca)^2\ge3abc(a+b+c).\]
Hint. Apply \(x^2+y^2+z^2\ge xy+yz+zx\) to \(x=ab\), \(y=bc\), \(z=ca\).
Let \(x=ab\), \(y=bc\), \(z=ca\). Then \((x+y+z)^2\ge3(xy+yz+zx)\). The left side is \(q^2\), and \(xy+yz+zx=abc(a+b+c)=pr\). Therefore \(q^2\ge3pr\).
An important basic relation among \(p,q,r\).