Hello hope you're having a good day
I need help in solving these problems
Find a set of real numbers (R) such as
sqrt(ab)+sqrt(cd) >= sqrt(a+b) + sqrt (c+d)
the Second is solving this
a and b are integers
This is a quadratic in a. Now take a look at the discriminant:
For an integer solution for a we must require that is a perfect square. Are there any integer b that do this? If you find solutions to b, use the quadratic equation in a,
to find if this gives an integer value for a.
Factor all perfect squares from the RHS:
That means that we need a value of an integer b such that is a perfect square. If you don't see it, start picking values for b and see what you get. Note that, since b is squared all you need to do is check all the non-negative values for b. (And, of course once you do this, you still need to verify that these solutions for b give an integer value for a.)
Hint: There are only two values of b that make a perfect square. Can you prove this?
First remove any perfect square factors from . So we require that must be a perfect square.
There are only two possibilities for b: b is even or it is odd.
If b is even then let b = 2n. This implies that . We need to be an even number divisible by 2 to make this a square number. But is odd. Thus b cannot be an even number.
So let b be an odd number: b = 2n + 1. Then . Now, 4 is a perfect square, so again I am going to factor that out of the problem. This leaves the statement that must be a square. By inspection n = 0 makes this a perfect square. So n = 0 implies b = 1. Further, the only other way that can be a perfect square is to let . This has the solution n = 1, which says that b = 3.
So we have only two possible solutions for an integer b: b = 1 or b = 3. Both of these give integer solutions for a.