if 3 positive real numbers x,y,z are in AP such that xyz=4,then what will be the minimum value of y?i)2^1/3 ii)2^2/3 iii) 2^1/4 iv) 2^3/4

let the numbers be a-d, a, a+d
So, xyz=4, this means (a-d)(a)(a+d)=4
=> $$a(a^{2}-d^{2})=4$$
=> $$a^{3}(1-\frac{d^{2}}{a^2})=4$$
We have to minimize a, this means we have to minimize $$a^{3}$$, this means we have to maximize $$(1-\frac{d^{2}}{a^2})$$. Since $$\frac{d^{2}}{a^2}$$ is always positive, $$(1-\frac{d^{2}}{a^2})$$ has max value of 1. This means minimum $$a^{3}=4$$. This means minimum $$a=2^{\frac{2}{3}}$$