### 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: This problem looks complicated, but I notice that it involves a lot of fractions involving $x$ and $y.$
I wonder if I can simplify it by making a substitution, like $u=\dfrac{x}{y}$ or $v=\dfrac{y}{x}.$
Let me try $u=\dfrac{x}{y}.$
Then I can rewrite the equation as \[3=k^2\left(u^2+\dfrac{1}{u^2}\right)+k\left(u+\dfrac{1}{u}\right).\]
This looks a bit nicer, because it only involves one variable, $u.$
Now, I want to find the maximum possible value of $k,$ so I need to think about how $k$ depends on $u.$
I can try to isolate $k$ by rearranging the equation, or I can look for patterns or symmetries in the equation.
I notice that the equation is symmetric in $u$ and $\dfrac{1}{u},$ meaning that if I swap them, the equation does not change.
This means that $u$ and $\dfrac{1}{u}$ must have the same effect on $k.$
This suggests that $k$ might reach its maximum value when $u$ and $\dfrac{1}{u}$ are equal, that is, when $u=1.$
Let me check if this is true by plugging in $u=1$ into the equation.
I get \[3=k^2(1^2+1^2)+k(1+1)=2k^2+2k.\]
To find the maximum value of $k,$ I can use the quadratic formula or complete the square.
I prefer the latter, because it gives me a nice expression for the vertex of the parabola.
I get \[3=2\left(k+\dfrac{1}{2}\right)^2-\dfrac{1}{2}.\]
Adding $\dfrac{1}{2}$ to both sides and dividing by $2,$ I get \[\left(k+\dfrac{1}{2}\right)^2=\dfrac{7}{4}.\]
Taking the square root and subtracting $\dfrac{1}{2},$ I get \[k=\pm\sqrt{\dfrac{7}{4}}-\dfrac{1}{2}.\]
However, I remember that $k$ has to be positive, so I ignore the negative root and take the positive one.
This gives me \[k=\sqrt{\dfrac{7}{4}}-\dfrac{1}{2}.\]
This is the maximum possible value of $k,$ because if I choose any other value of $u,$ the value of $k$ will decrease.
I can check this by taking the derivative of the expression for $k$ in terms of $u,$ but I won't do that here.