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Let $x,y,z\in [1,4]$ such that $x \geq y$ and $x \geq z$.

Find the minimum value of this expression: $ P=\frac{x}{2x+3y}+\frac{y}{y+z}+\frac{z}{z+x} $

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    or you can transform into polar coordinates, reduce the variables to two(same as DKhanh does), then apply [second partial derivative test](http://en.wikipedia.org/wiki/Second_partial_derivative_test)2011-08-26

2 Answers 2

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As has been mentioned in comments, we want to find the minimum value of $ P=\frac{1}{2+3u}+\frac{u}{u+v}+\frac{v}{v+1} $ for $u,v\in[\frac{1}{4},1]$. Take partials of $P$ with respect to $u$ and $v$: $ \begin{align} \frac{\partial P}{\partial u}&=-\frac{3}{(2+3u)^2}+\frac{v}{(u+v)^2}\\ \frac{\partial P}{\partial v}&=-\frac{u}{(u+v)^2}+\frac{1}{(v+1)^2} \end{align} $ To find an interior extremum, we need $\frac{\partial P}{\partial u}=\frac{\partial P}{\partial v}=0$. In that case, we need $ 0=u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}=-\frac{3u}{(2+3u)^2}+\frac{v}{(v+1)^2} $ However, for $u,v\in[\frac{1}{4},1]$, we have $ \begin{array}{c} \frac{12}{121}\le\frac{3u}{(2+3u)^2}\le\frac{3}{25}\\ \frac{4}{25}\le\frac{v}{(v+1)^2}\le\frac{1}{4} \end{array} $ Thus, $u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}\ge\frac{1}{25}$, so there can be no interior extremum.

Because $(u,v)\cdot\nabla P=u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}\ge\frac{1}{25}$ everywhere in $[\frac{1}{4},1]\times[\frac{1}{4},1]$, the minimum must be taken on the left or bottom edge of that square; i.e. $u=\frac{1}{4}$ or $v=\frac{1}{4}$.


$u=\frac{1}{4}$: $\frac{\partial P}{\partial v}=-\frac{4}{(1+4v)^2}+\frac{1}{(v+1)^2}=\frac{12v^2-3}{(1+4v)^2(v+1)^2}$ which vanishes at $v=\frac{1}{2}$.

Because $P(\frac{1}{4},\frac{1}{4})=\frac{117}{110}$ and $P(\frac{1}{4},\frac{1}{2})=\frac{34}{33}$ and $P(\frac{1}{4},1)=\frac{117}{110}$, $P(\frac{1}{4},\frac{1}{2})$ is a local minimum.


$v=\frac{1}{4}$: $\frac{\partial P}{\partial u}=-\frac{3}{(2+3u)^2}+\frac{4}{(4u+1)^2}=\frac{-12u^2+24u+13}{(2+3u)^2(4u+1)^2}$ which does not vanish on $[\frac{1}{4},1]$.

Because $P(\frac{1}{4},\frac{1}{4})=\frac{117}{110}$ and $P(1,\frac{1}{4})=\frac{33}{20}$, $P(\frac{1}{4},\frac{1}{4})$ is a local minimum.


Thus, $P(\frac{1}{4},\frac{1}{2})=\frac{34}{33}$ is the minimum of P for $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$.

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    @DKhanh: Following Sasha's comment, $u = y/x$, $v = z/x$. Since $y\le x$ and $x,y\in[1,4]$, we have that $u\in[\frac{1}{4},1]$. Since $z\le x$ and $x,z\in[1,4]$, we have that $v\in[\frac{1}{4},1]$. I used that $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$.2011-08-27
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The answer is $\frac{34}{33}$.

To see this, start with the fact that the derivative $\partial_zP(x,y,z)$ of $P(x,y,z)$ with respect to $z$ has the sign of $(x-y)(z^2-xy)$.

  1. Assume that $x>y$.

    Then $\partial_zP<0$ at $z=1$ hence $P$ is not minimal at $(x,y,1)$ and $\partial_zP>0$ at $z=x$ hence $P$ is not minimal at $(x,y,x)$. Thus $z^2=xy$.
    Define $Q$ by $Q(w)=\dfrac1{2+3w^2}+\dfrac{2w}{1+w}$, then $P(x,y,\sqrt{xy})=Q(u)$ with $u=\sqrt{y/x}$ hence $u\in[\frac12,1]$. Now, Q''>0 on the interval $[\frac12,1]$ and Q'(\frac12)>0 hence Q'>0 on $[\frac12,1]$. Thus, $Q(u)\ge Q(\frac12)$ for every $u$ in $[\frac12,1]$.

  2. Assume that $x=y$.

    Then $P(x,x,z)=1+\frac15$ for every $z$.

Finally $Q(\frac12)=1+\frac1{33}<1+\frac15$ hence $P$ is minimum at $(4,1,2)$ where its value is $P(4,1,2)=Q(\frac12)=1+\frac1{33}$.