Mock AIME II 2012 Problems/Problem 7
Given are positive real numbers that satisfy , then the value can be expressed as , where and are relatively prime positive integers. Find .
Proof: We can rearrange the given equation to get . Notice that the RHS is almost a perfect square. . Notice now that the RHS is always positive, and so we must have . Since is positive, we can divide by to get .
Seeing that we can make perfect squares with the linear terms and square root terms, we try to force perfect squares. Since the coefficient of is already a perfect square, we begin with that. We have . Now we divide both sides of the equation by to get . We recognize from our original equation as , and we substitute that back in to get . We again recognize a template for a perfect square in the right hand fraction, so we have . We can also right the other terms in the numerator of the right hand fraction as perfect squares in terms of and . . We now see that . We can take the square root of both sides to see the full picture. . This equation is just the equality case for the RMS-AM inequality! Therefore, from the equality case, we must have , from which we find that and . We see that these satisfy the original equality, and so their product is .
Note: We had to prove that so that , one of the terms in the RMS-AM inequality, is positive, otherwise the inequality is false.
Let and . We have . Moving this to one side, we have . Completing the square, we have
The above simplifies into , and since squares are nonnegative, we have both squares equal to , which means and . Thus, and , so and .