# How to find the minimum value of $\frac{x}{2x+3y}+\frac{y}{y+z}+\frac{z}{z+x}$?

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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Since $x \not= 0$, denote $y= u x$ and $z=v x$. You are, then, looking for minimum in $\frac{1}{4} \le u \le 1$ and $\frac{1}{4} \le v \le 1$. – Sasha Aug 26 '11 at 6:06
@Sasha: following to your suggestion, we get $P=\frac{1}{2+3u}+\frac{u}{u+v}+\frac{v}{v+1}$ What then shall we do? – KevinBui Aug 26 '11 at 9:48
or you can transform into polar coordinates, reduce the variables to two(same as DKhanh does), then apply second partial derivative test – newbie Aug 26 '11 at 10:54

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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I just noticed the algebra-precalculus tag. Using partials might not be good. – robjohn Aug 26 '11 at 15:02
why don't you apply the condition $x>y , x>z$? – KevinBui Aug 27 '11 at 2:29
@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 $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$. – robjohn Aug 27 '11 at 3:58

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 $P(4,1,2)=Q(\frac12)=1+\frac1{33}$.

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