Using Lagranges mean value theorem, prove that `(b-a)/bltlog(b/a)lt(b-a)/a `,where `0ltaltb`

Explanation

Lagranges Mean value theorem states that if a function `f(x)` is continuous and differentiable in interval `(a,b)`, then,

`(f(b)-f(a))/(b-a) = f'(c)`, where c lies in `(a,b)`.

Let `f(x) = logx` where `x in [a,b]` such that `0 lt a lt b`


Then, from Lagranges Mean value theorem,

` f'(c) = (f(b)-f(a))/(b-a) `

`1/c = (logb-loga)/(b-a)->(1)`

Here, `a lt c lt b`

...