Solve the inequality x^square root x=

Please consider this answer though
late:

To solve for x^(sqrtx) =<
(sqrtx)^x.

We square both sides x^(2x^1/2) =<
{(sqrtx)^x}^2

=> x^(2x^1/2) ={
x(1/2)}2x

=> x^(2x^1/2) =<
x^x.

There are 3 cases when the above inequality holds.:
x< 1 , x= 1 and x > 1. We compare the powers of both sides in 3
cases:

The inequality holds under the following
situations:

2x^1/2 > or= or < x,
according as x < 1 or x = 1 or x>
1.

=> 4x > or = or < x^2 according as
x <1 or x= 1or x> 1.

x^2-4x < 0. Or
x(x-4) < 0 when 0 < x <1. This gives x<4 and x< 1.
So 0 < x <1, the inequality
holds.

x^2-4x = 0 when x = 0. This gives
x= 0, the equality
holds.

x^2-4x >0. Or x(x-4) >
0 when x>1. This gives x > 4 the inequality
holds.

Check:

When
0 < x< 1:  Put x = 1/9 in x^square root x=<(square root x)^x. LHS =
(1/16) sqrt(1/16)) = (1/16)^(1/4) = 1/2. RHS = {sqrt(1/16)}^(1/16) = (1/4)^1/16) =
0.917. So LHS < RHS.

When x = 1,x^square root
x=<(square root x)^x becomes 1^(sqrt1) = (sqrt1)^1
obviously.

When x > 4,  we put x= 16 in x^square
root x=<(square root x)^x and get LHS 16^(sqrt16)  = 16^4 = 4^8. RHS =
(sqrt16)^16 = 4^16. So LHS =< RHS.

So
the given inequality x^square root x < (square root x)^x holds when  0 <
x< 1 or when x > 4 and the equality, x^square root x = (square root x)^x
holds when x= 1.