SOLUTION: Show that 1/x <= 3 - 2sqrt(x) for all positive real numbers x. Describe when we have equality.

Algebra ->  Inequalities -> SOLUTION: Show that 1/x <= 3 - 2sqrt(x) for all positive real numbers x. Describe when we have equality.      Log On


   



Question 1178757: Show that 1/x <= 3 - 2sqrt(x) for all positive real numbers x. Describe when we have equality.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
1%2Fx+%3C=+3+-+2sqrt%28x%29
Since x is positive we can multiply through by x without
changing the inequality sign.

1+%3C=+3x+-+2x%2Asqrt%28x%29

2x%2Asqrt%28x%29%3C=+3x+-+1

%282x%2Asqrt%28x%29%29%5E2%3C=+%283x+-+1%29%5E2

4x%5E2%2Ax+%3C=+9x%5E2-6x%2B1

4x%5E3+%3C=+9x%5E2-6x%2B1

4x%5E3-9x%5E2%2B6x-1%3C=0

1 | 4  -9   6  -1
  |     4  -5   1
    4  -5   1   0

%28x-1%29%284x%5E2-5x%2B1%29%3C=0

%28x-1%29%28x-1%29%284x-1%29%3C=0

%28x-1%29%5E2%284x-1%29%3C=0

The critical numbers are 1 and 1/4

The solution is (0,1/4) U {1}

However the interval (0,1/4) is ruled out by substituting
the test value x = 0.1 from that interval in the original
inequality:

1%2Fx+%3C=+3+-+2sqrt%28x%29

1%2F0.1+%3C=+3+-+2sqrt%280.1%29

10+%3C=+3-2%280.316227766%29

10%3C=2.367544468

which is clearly false.

Thus the only solution is {1} and that is when equality holds.

In other words the inequality

1%2Fx+%3C=+3+-+2sqrt%28x%29

is equivalent to the equation:

1%2Fx+=+3+-+2sqrt%28x%29

because the left side can NEVER be less than the right side.

So we can ONLY and ALWAYS have equality, and x=1 is the ONLY
value x can take on

Edwin