Let p ∈ Q be a rational number such that p > 1 and letq := pp − 1 .Prove that for all a, b ≥ 0,ab ≤ app + bqq

To prove that for all a, b ≥ 0, ab ≤ app + bqq, we can start by rearranging the inequality.

First, let's substitute q with pp - 1:

ab ≤ app + b(pp - 1)(pp - 1)

Next, we can expand the right side of the inequality:

ab ≤ app + b(p^2p^2 - 2pp + 1)

Simplifying further:

ab ≤ app + bp^4 - 2bp^3 + b

Now, let's factor out p from the second and third terms:

ab ≤ app + bp^3(p - 2) + b

Since p > 1, we know that p - 2 < 0. Therefore, we can replace p - 2 with -(2 - p):

ab ≤ app + b(-(2 - p)p^3) + b

Simplifying again:

ab ≤ app - b(2 - p)p^3 + b

Now, let's factor out b from the second and third terms:

ab ≤ app - b(2 - p)p^3 + b

ab ≤ app + bp(2 - p)p^2

Finally, we can rewrite the right side of the inequality as a product:

ab ≤ app + b(p^2)(2 - p)p

Now, we can see that the right side of the inequality is a product of three non-negative terms: app, p^2, and (2 - p)p. Since a and b are also non-negative, the product of these terms will always be greater than or equal to ab.

Therefore, we have proven that for all a, b ≥ 0, ab ≤ app + bqq.