# Show 3^root 5<=5^root 3 for f =ln x /root x

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You need to take logarithms both sides of inequality, such that:

`ln 3*sqrt 5 <= ln 5*sqrt 3`

You should re-arrange the members of inequality such that:

`ln 3/sqrt 3 <= ln 5/sqrt 5`

You should notice that the terms of inequality `ln 3/sqrt 3` and ` ln 5/sqrt 5` represent the values of the function at `x = 3` and `x = 5` .

You need to test the monotony of the function over interval `(3,5)` , hence, you should perform derivative test, such that:

`f'(x) = 0 => ((ln x)'*sqrt x - ln x*(sqrt x)')/x = 0`

`(sqrt x/x - ln x/(2sqrt x))/x = 0 => sqrt x/x - ln x/(2sqrt x) = 0`

`2sqrt x - ln x*sqrt x = 0 => sqrt x(2 - ln x) = 0`

Using the zero product property yields:

`sqrt x = 0 => x = 0`

`2 - ln x = 0 => ln x = 2 => x = e^2`

You should notice that the derivative `f'(x) = (sqrt x/x - ln x/(2sqrt x))/x` is positive over the interval `(0,e^2)` , hence, the function increases over `(0,e^2).`

Since `(3,5) sub (0,e^2)` , hence the function increases over `(3,5)` , such that:

`3 < 5 => (ln 3)/sqrt 3 < ln 5/sqrt 5`

**Hence, testing if the inequality `3^(sqrt5)<5^(sqrt 3)` holds over the interval `(3,5)` , using derivative test, yields that `(ln 3)/sqrt 3 < ln 5/sqrt 5` , hence, the inequality is valid.**