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| Math Central | Quandaries & Queries |
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Question from jerry, a student: Two real numbers a and b satisfy the equation ab=1. i) Prove that a^6 + 4b^6 >4 ii) Find out whether the inequality a^6 + 4b^6 >4 holds for all a and b such that ab = 1 |
Hi Jerry.
The two parts of the question seem the same to me.
Here's how I would solve it.
Let c = a6. Then 1/c = 1/a6 = b6. Then we are to show that:
c + 4/c > 4 for all positive real values of c.
Since c is positive, we can multiply both sides by c, then see a quadratic:
c2 + 4 > 4c
c2 - 4c + 4 > 0
which is a perfect square:
(c - 2)2 > 0
So the value of c = 2 is the only problem here.
Can you complete the question now?
Stephen La Rocque
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