Finding the integer solutions of the equation $3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14$

Question

$ 3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14 $ .

I already solved this using the Cauchy–Schwarz inequality and got $x=4$ and $y=5$. But I'm sure there is a prettier, simpler solution to this and I was wondering if anyone could suggest one.

Answer 1

The Cauchy-Schwarz inequality goes like this:

$LHS \leq \sqrt{3^2+2^2+1^2}\cdot \sqrt{\left(\sqrt{x+y}\right)^2+\left(\sqrt{8-x}\right)^2+\left(\sqrt{6-y}\right)^2}=14=RHS$.

Thus you have equality when $\dfrac{3}{\sqrt{x+y}}=\dfrac{2}{\sqrt{8-x}}=\dfrac{1}{\sqrt{6-y}} \to x=4,y=5$.

Answer 2

Based on what you wrote, we can deduct the following: $$x+y\geq 0$$ $$8-x\geq 0 $$ $$6-y \geq 0$$ As a result, $$14 \geq x+y\geq 0$$ You also know that $x+y$ is a square. Therefore, you have only 4 squares that satisfy this double inequality: $$ 0, 1, 4, 9$ You can quickly eliminate 0, 1 as options because it would to 2 irrationals.

As far as $4$, that'd generate $ (0,4),(4,0),(1,3),(3,1) $ and $(2,2)$ . Clearly none of them satisfies your equation.

$9$ generates $(3,6),(6,3),(4,5)$and$ (5,4)$. We cannot take into consideration $(0,9),(9,0)$ because $x\leq8$ and $y \leq 6$.

From here, given the fact that $8-x$ and $6-y$ must be perfect squares, it can be deducted the only couple that satisfies your equation is $(4,5)$.

Answer 3

This is a trivial solution so that we get answer in integer

8 - x and 6 - y are under the root so x <8 and y <6 , for making 6 - y to be perfect square y=2 or 5. and similarly x=4 or 7.

Now x+y is also under the root so x+y has to be perfect square. So we get two pairs of (x, y)=(7,2) and (2,5) but (7,2) will not satisfy the equation so ur answer is x=4, y=5.