For two positive real numbers a and b such that 1/a^2+1/b^3=4 ,the minimum value of the

Question

Answer 1

General term f * (a * x ^ (1/8) + b * x ^ (- 1/12)) ^ 10 * i*epsilon = ^ 10 C_{r} * (a * x ^ (1/8)) ^ (10 - r) * (b * x ^ (- 1/12)) ^ r On solving r = 6

= ( = ( a^ 4 b^ 6 )^ 1 4 >= 4/(2(1/(a ^ 2) + 1/(b ^ 2))) Rightarrow a^ 4 b^ 6 >=0 ^ 10 C_{6} * (1/2) ^ 4 = 210 16 Rightarrow 105 8