SOLUTION: If x is a real number, then show that 0<(1/x^2+6x+10)&#8804;1.

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Question 345494: If x is a real number, then show that 0<(1/x^2+6x+10)≤1.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
In order for 1%2F%28x%5E2%2B6x%2B10%29%3C=1, then x%5E2%2B6x%2B10%3E=1 for all x.
Convert to vertex form by completing the square, y=a%28x-h%29%5E2%2Bk where (h,k) is the vertex.
f%28x%29=x%5E2%2B6x%2B10
f%28x%29=%28x%5E2%2B6x%2B9%29%2B10-9
f%28x%29=%28x%2B3%29%5E2%2B1
Since the coefficient of x%5E2 is positive, the parabola opens upwards and the value at the vertex is the function minimum.
(h,k)=(-3,1)
Since the function value is never less than 1, then the inverse of the function value is never greater than 1.