### Instruction: If $x,$ $y,$ and $k$ are positive real numbers such that \[3=k^2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+k\left(\dfrac{x}{y}+\dfrac{y}{x}\right),\]find the maximum possible value of $k.$.### Response: I notice that this equation has some symmetry in $x$ and $y,$ and it involves squares and reciprocals of $x$ and $y.$
This makes me think of using the arithmetic mean-geometric mean inequality, or AM-GM, which says that for any nonnegative numbers $a$ and $b,$ we have \[\dfrac{a+b}{2}\geq\sqrt{ab}.\]
I also notice that the equation has two terms on the right-hand side, one involving $k^2$ and the other involving $k.$
This makes me think of completing the square, which is a technique for solving quadratic equations by adding and subtracting a constant term to make a perfect square.
So I try to combine these two ideas and see what happens.
First, I rewrite the equation as \[3=k^2\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-2\right)+k\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)+2k^2+2k.\]
Then, I apply AM-GM to each of the parentheses, and get \[3\geq k^2\left(2\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y^2}{x^2}}-2\right)+k\left(2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}-2\right)+2k^2+2k.\]
Simplifying the radicals, I get \[3\geq k^2(2-2)+k(2-2)+2k^2+2k.\]
This reduces to \[3\geq 2k^2+2k.\]
Dividing by 2, I get \[1.5\geq k^2+k.\]
Now, I complete the square by adding and subtracting $(0.5)^2$ to the right-hand side, and get \[1.5\geq k^2+k+(0.5)^2-(0.5)^2.\]
This gives me \[1.5\geq (k+0.5)^2-0.25.\]
Adding 0.25 to both sides, I get \[1.75\geq (k+0.5)^2.\]
Taking the square root of both sides, I get \[\sqrt{1.75}\geq |k+0.5|.\]
This means that \[k+0.5\leq\sqrt{1.75}\] and \[k+0.5\geq -\sqrt{1.75}.\]
Subtracting 0.5 from both sides, I get \[k\leq\sqrt{1.75}-0.5\] and \[k\geq -\sqrt{1.75}-0.5.\]
Since $k$ is positive, I can ignore the second inequality and focus on the first one.
This tells me that the maximum possible value of $k$ is \[\sqrt{1.75}-0.5.\]