# Solve the inequality x^square root x=<(square root x)^x

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### 3 Answers

We have to solve: x^sqrt x =< (sqrt x)^x

x^sqrt x =< (sqrt x)^x

subtract (sqrt x)^x from both the sides

=> x^sqrt x - (sqrt x)^x =< 0

For x = 4, 4^2 = 16 and 2^4 = 16

and x^sqrt x - (sqrt x)^x = 0

For values of x above 4 we see that x^sqrt x - (sqrt x)^x < 0

For values of x below 4 we see that x^sqrt x - (sqrt x)^x > 0

The inequality x^sqrt x - (sqrt x)^x =< 0 holds for values of x =< 4

=> x^sqrt x =< (sqrt x)^x for x =< 4

Also, at x = 1, we see that x^sqrt x = (sqrt x)^x = 1

**The values of x lie in [4 , +.inf} U {1}**

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 hold**s.

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.**

We'll take natural logarithms both sides, to keep the sense of inequality and to use the power rule of logarithms:

ln (x^sqrtx) =< ln (sqrt x)^x

sqrt x* ln x =< x* ln (sqrt x)

sqrt x* ln x =< x* ln (x)^(1/2)

sqrt x* ln x =< (x/2)* ln (x)

We'll divide by ln x:

sqrt x =< (x/2)

2*sqrt x =< x

If x belongs to the interval (0,1), then the inequality 2*sqrt x =< x <=> 4x =< x^2 is verified.

If x > 1, then the inequality sqrt x =< x/2 is also verified.

**Therefore, the interval of admissible solutions for the inequality (x^sqrtx) =< (sqrt x)^x to hold is (0 , infinite).**