Question 1178757
{{{1/x <= 3 - 2sqrt(x)}}}
<pre>
Since x is positive we can multiply through by x without
changing the inequality sign.

{{{1 <= 3x - 2x*sqrt(x)}}}

{{{2x*sqrt(x)<= 3x - 1}}}

{{{(2x*sqrt(x))^2<= (3x - 1)^2}}}

{{{4x^2*x <= 9x^2-6x+1}}}

{{{4x^3 <= 9x^2-6x+1}}}

{{{4x^3-9x^2+6x-1<=0}}}

1 | 4  -9   6  -1
  |<u>     4  -5   1</u>
    4  -5   1   0

{{{(x-1)(4x^2-5x+1)<=0}}}

{{{(x-1)(x-1)(4x-1)<=0}}}

{{{(x-1)^2(4x-1)<=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/x <= 3 - 2sqrt(x)}}}

{{{1/0.1 <= 3 - 2sqrt(0.1)}}}

{{{10 <= 3-2(0.316227766)}}}

{{{10<=2.367544468}}}

which is clearly false.

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

In other words the inequality

{{{1/x <= 3 - 2sqrt(x)}}}

is equivalent to the equation:

{{{1/x = 3 - 2sqrt(x)}}}

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</pre>