Problem
ALG-B2-M03-P008 Fixed denominator sum
#8
★★★★☆ Level 4 of 5
Let \(a,b,c>0\), \(a+b+c=6\). Find the minimum of \(\frac{4}{a}+\frac{9}{b}+\frac{16}{c}\).
Hint 1. Write the numerators as \(2^2,3^2,4^2\).
Hint 2. Cauchy gives \(\frac{(2+3+4)^2}{a+b+c}\).
By Cauchy, \[\frac{4}{a}+\frac{9}{b}+\frac{16}{c}\ge\frac{(2+3+4)^2}{a+b+c}=\frac{81}{6}=\frac{27}{2}.\] Equality occurs when \(\frac{2}{a}=\frac{3}{b}=\frac{4}{c}\), i.e. \(a:b:c=2:3:4\). This is attainable with sum \(6\), so the minimum is \(\frac{27}{2}\).
Cauchy-Schwarz module training problem. Method tags: cauchy, equality-case.