Math CentralQuandaries & Queries


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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