Exams › JEE Advanced › Maths
Correct answer: 96
Expression = (x+2y)² + 2z². Let s = x+2y. By AM-GM: x+2y >= 2*sqrt(2xy), so (x+2y)² >= 4*2*xy = 8xy. Thus expression >= 8xy + 2z². Now minimize 8xy + 2z² with xyz=32. By AM-GM on 8xy, 8xy, 2z²: (8xy+8xy+2z²)/3 >= (8xy*8xy*2z²)^(1/3). At minimum, 8xy = 2z², so z² = 4xy, z=2*sqrt(xy). Also xyz=32: xy*2*sqrt(xy)=32 => 2(xy)^(3/2)=32 => (xy)^(3/2)=16 => xy=16^(2/3). Hmm, let me try: set a=x, b=2y, so (a+b)²+2z², with xyz=32 => a*(b/2)*z=32 => abz=64. Minimize (a+b)²+2z² subject to abz=64. AM-GM: a+b>=2*sqrt(ab). Min of (a+b)² is 4ab when a=b. So minimize 4ab+2z² with abz=64. Set 4ab=2z² (by AM-GM for minimum), so z²=2ab, z=sqrt(2ab). Then ab*sqrt(2ab)=64 => (2ab)^(3/2)/sqrt(2)=64 => (2ab)^(3/2)=64*sqrt(2) => 2ab=16 => ab=8. Then z=4. Total=4*8+2*16=32+32=64. But min of (a+b)²=4ab=32, so total=32+32=64. So answer is 64.