Let $x$ and $y$ are real numbers such that $\frac{1}{2}<\frac{x}{y}<2$. Find the minimum value $\begin{align*} \frac{x}{2y-x}+\frac{2y}{2x-y} \end{align*}$.
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1Welcome to stackexchange. This is the second question you've asked in the last half hour that shows no work of your own. When you post here you should include in your question what you tried and where you are stuck. You seem to have gotten answers anyway, but often you won't. – Ethan Bolker Mar 03 '18 at 16:57
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I'm a new member here, and I do not understand the rules here. I apologize for my mistake. – Yanmath Mar 03 '18 at 17:01
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Can you write the expression you are trying to bound in terms of $z=\frac xy$? – Mark Bennet Mar 03 '18 at 17:02
3 Answers
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$$\frac{x}{2y-x}+\frac{2y}{2x-y}=\frac{x/y}{2-x/y}+\frac{2}{2x/y-1}=\frac{z}{2-z}+\frac{2}{2z-1}$$
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 $2.8856181$ )
Mohammad Riazi-Kermani
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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
Dr. Sonnhard Graubner
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4@Yanmath No, you show your work please. This isn't a do-my-homework site and you've been given a pretty strong hint. – Mar 03 '18 at 16:59
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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.
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