Problem
ALG-B2-M09-P017 Product with fixed q
#17
★★★★★ Level 5 of 5
Let \(a,b,c>0\), \(ab+bc+ca=3\). Prove \(abc\le1\).
Hint. Use \(q^2\ge3pr\) and \(p^2\ge3q\).
Let \(p=a+b+c\), \(q=3\), \(r=abc\). From \(p^2\ge3q=9\), we get \(p\ge3\). Also \(q^2\ge3pr\), i.e. \(9\ge3pr\), so \(r\le3/p\le1\). Therefore \(abc\le1\). Equality holds at \(a=b=c=1\).
A good problem combining two \(p,q,r\) relations.