Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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
up vote 4 down vote accepted

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 $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$.

share|cite|improve this answer
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 $P(4,1,2)=Q(\frac12)=1+\frac1{33}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.