Please help me to solve this problem

Question

Let $x$ and $y$ are real numbers such that $\frac{1}{2}<\frac{x}{y}<2$. Find the minimum value $\begin{align*} \frac{x}{2y-x}+\frac{2y}{2x-y} \end{align*}$.

Answer 1

it is $$\frac{x}{2y-x}+\frac{2y}{2x-y}\geq \frac{1}{3}(3+4\sqrt{2})$$ define $$f(x,y)=\frac{x}{2y-x}+\frac{2y}{2x-y}$$ and compute the partial derivatives

Answer 2

$$\frac{x}{2y-x}+\frac{2y}{2x-y}=\frac{x/y}{2-x/y}+\frac{2}{2x/y-1}=\frac{z}{2-z}+\frac{2}{2z-1}$$

where $ \frac{1}{2}<z<2$

Note that $$g(z) = \frac{z}{2-z}+\frac{2}{2z-1}$$ is a function of just one variable $z$

You can find the minimum of $g(z)$ on $ \frac{1}{2}<z<2$ by differentiation.( My answer was $2.8856181$ )

Answer 3

Let $t=\dfrac xy$. Then $$f=\frac{x}{2y-x}+\frac{2y}{2x-y}=\frac{t}{2-t}+\frac2{2t-1}=2\left(\frac{1}{2-t}+\frac1{2t-1}\right)-1$$ Differentiating gives $$\frac{df}{dt}=2\left(\frac1{(2-t)^2}-\frac2{(2t-1)^2}\right)=0$$ for stationary points so $$(2t-1)^2=2(2-t)^2\implies 4t^2-4t+1=2t^2-8t+8\implies 2t^2+4t-7=0$$ Now find the root between $\dfrac12$ and $2$, plug it back into the original expression and prove that its second derivative is greater than $0$ to show that it is a minimum.