Solve Challenging Math Problems: Prove Inequalities and Modulus Equations

[123apple]

Hi guys,

I'm going to a math competition soon, and there are a few problems that have stumped both me and my teacher. I was wondering if any of you could help me out?

Prove

(a^2+b^2)/(a+b) >= sqrt(ab)

for all real A and B.


p(x)mod(x-19) = 99
p(x)mod(x-99) = 19
p(x)mod(x-19)(x-99) = ?


Suppose for an integer, n, there exists numbers 2n+1 and 3n+1 that are perfect squares. Prove that n cannot be prime.



Thanks a ton for the help! I appreciate it.

 

[StatusX]

For the first, play around a little. In these kinds of problems, you should always start by trying to form squares, since if c-d is a square, it's pretty clear c>=d.

For the third, what is the difference of two squares?

 

[Fermat]

For the first question, the statement of your problem is inaccurate. It certainly isn't true for a and b of opposite sign, since you can't take the square root of a negative number. And ab > 0 is crucial to the proof.

Start by squaring both sides of the inequality, and multiplying up by (a+b)², giving (a²+b²)² >= ab(a+b)²
Evaluate the terms in the inequality and simplify.
Use the substitution a = b+k.
Re-evaluate the terms and simplify.
Take all terms over to the lhs.
You should now be able to show that all the terms on the lhs are always positive, thus satisfying the assumption that the inequality is true.

 

[123apple]

Sorry guys. For the second, it is all postive reals, not all reals. (I was reciting these from memory.)

Fermat, how do I know that the lhs is always positive?

I came up with this:

b^4 + (5k+1)b^3 + (6k^2 +3k)b^2 + (4k^3 + 3k^2)b + (k^4 +k^3) >= 0

If k is negative, how does this work out?



For the third one, the difference of two squares n and b is n^2 - b^2 = (n-b)(n+b)
I really don't see the connection to that problem.

Thanks for the help guys!

 

[Fermat]

I don't get the same expression as yourself. Check your working. In problems like this, I check my working as follows.

You should have, at one point, have gotten,

(a² + b²)² - ab(a + b)² >= 0

Now, this is (like) an identity, true for all (valid) values of a and b (at least that is what we are trying to prove, so for the moment we are going to assume it is true)
Since it is true for all values, then let a = 2, b = 1 and k = 1 (a = b+k).
Substituing these values into the expression for the inequality, that expression evaluates to 7 >= 0. No matter how we expand that expression and manipulate it, it will always evaluate to the value 7 (provided we don't multiply or divide).
So if you start from the inequality I have above, and expand it, simplify, substitute, etc, and check the evaluation of the expression every few lines, you will catch where any arithmetical errors occur.

Oh, and it doesn't matter if k is negative since you will (eventually) end up with even powers of it.