SOLUTION: If a, b, and c are positive constants, prove that {{{ax + b/x >= c}}} for all positive values of x when {{{4ab >= c^2}}}.

Algebra ->  Functions -> SOLUTION: If a, b, and c are positive constants, prove that {{{ax + b/x >= c}}} for all positive values of x when {{{4ab >= c^2}}}.       Log On


   

Question 1183866: If a, b, and c are positive constants, prove that ax+%2B+b%2Fx+%3E=+c for all positive values of x when 4ab+%3E=+c%5E2.
Found 2 solutions by robertb, Edwin McCravy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
*
highlight%28NOTE%29: As a response to tutor Edwin McCravy's response, I looked at the transcript again, and have inserted the phrase "is floating
on or" before the "tangent" word, which was actually what I originally had in mind, but somehow came out different (probably due to fast typing).
Otherwise (if the parabola is always tangent to the x-axis) I will just be saying that the discriminant will just be equal to 0,
that is c%5E2+-+4ab+=+0which is quite far from the result we wanted to prove.
But thanks for detecting that!
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If ax+%2B+b%2Fx+%3E=+c for all positive values of x, then
ax%5E2+%2B+b+%3E=+cx <===> ax%5E2+-cx+%2B+b+%3E=+0.

This is a quadratic equation whose graph (a parabola) is opening upward and is floating on or tangent to the x-axis.
This means that the discriminant of the function must be less than or equal to 0 or

c%5E2+-+4ab+%3C=+0 ===> highlight%28c%5E2+%3C=+4ab%29.

And the statement is PROVED.

Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!
I agree with everything robertb said, except 

"tangent to the x-axis"

It isn't tangent to the x-axis.  It is ABOVE the x axis, tho.

Substitute a=b=c=1 and x2-x+1 ≥ 0.  That quadratic is
ABOVE, but not tangent to, the x-axis.

Edwin