Question 37Given two positive numbers $a,$ $b$ such that $a\lt b.$ Let A.M. be their arithmetic mean and let G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than:$\dfrac{(b+a)^2}{ab}$$\dfrac{(b+a)^2}{8b}$$\dfrac{(b-a)^2}{ab}$$\dfrac{(b-a)^2}{8a}$$\dfrac{(b-a)^2}{8b}$