Problem
ALG-B2-M03-P012 Minimum of a fractional sum
#12
★★★★★ Level 5 of 5
Let \(x,y,z>0\) and \(x+y+z=10\). Find the minimum of \(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\).
Hint 1. The numerators are \(1^2,2^2,3^2\).
Hint 2. Check equality attainability.
By Cauchy, \[\frac1x+\frac4y+\frac9z\ge\frac{(1+2+3)^2}{x+y+z}=\frac{36}{10}=\frac{18}{5}.\] Equality occurs when \(1/x=2/y=3/z\), i.e. \(x:y:z=1:2:3\). This is attainable with sum \(10\), so the minimum is \(\frac{18}{5}\).
Cauchy-Schwarz module training problem. Method tags: cauchy, equality-case.