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Let $x,y,z>0$ and $x+y+z=1$, then find the least value of $${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$

I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.

I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.

Dhanush Krishna
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5 Answers5

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Here we use the AM-HM (harmonic mean) inequality. \begin{align*} {{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}} &=-3+2\left(\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}\right)\\ &\geq -3+2\left[\frac{3^2}{(2-x)+(2-y)+(2-z)}\right]\\ &=-3+\frac{18}{5}=\frac{3}{5}. \end{align*} Also, when $x=y=z=\frac{1}{3}$, the above equality holds. Therefore the least value is $\frac{3}{5}$.

Solumilkyu
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    Good +1, though that inequality is more recognized as the Engel form of the Cauchy Schwarz inequality – Macavity Mar 22 '16 at 08:08
  • Very nice (+1), treating $$\left(\frac1{2-x}+\frac1{2-y}+\frac1{2-z}\right)$$ as (three times) the arithmetic mean! – bgins Mar 22 '16 at 09:50
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You can also use Cauchy-Schwarz: $$\left(x(2-x)+y(2-y)+z(2-z)\right)\left(\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\right)\ge (x+y+z)^2=1.$$ Furthermore, $$\begin{aligned}x(2-x)+y(2-y)+z(2-z) &= 2(x+y+z)-(x^2+y^2+z^2)\\ &\le 2-\frac{(x+y+z)^2}{3}=\frac53. \end{aligned}$$ Thus the minimum is $3/5$ attained at $x=y=z=1/3$.

Quang Hoang
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For the sake of alternatives:

It is easy to check that $$\frac{x}{2-x} \ge \frac{18x-1}{25} \quad\forall x < 2,$$ (equivalent to $(3x-1)^2\ge 0$). Applying the above inequality for $y$ and $z$ also, then taking the sum of the three inequalities, we are done.

f10w
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Mike Yu's answer is excellent. For the sake of alternatives, I present a solution using the method of Lagrange Multipliers. Because of the symmetry in the forms of both functions, we can intuitively see the solution.

The function $h(x)=\frac{x}{2-x}$ decreases from (a limit or pole at) positive infinity to $0$ on the interval $x\in(0,2]$. On this interval, the larger $x$ is, the smaller $h(x)$ will be ($h$ is strictly decreasing). Each of $x,y,z$ need to be large to minimize the sum. By symmetry, the solution must be when each variable is equal: $x=y=z=\tfrac13$, so that the sum is $3h(\tfrac13)=\tfrac35$.

But to justify this symmetry argument, we might need to use Lagrange Multipliers. Taking the gradient (directional derivative) of the objective and constraint functions $$ \begin{align} f(x,y,z)&={x\over2-x}+{y\over2-y}+{z\over2-z}\\ g(x,y,z)&=x+y+z-1 \end{align} $$ we get vector valued functions $\nabla f=(f_x,f_y,f_z)=\left({\partial f\over\partial x},{\partial f\over\partial y},{\partial f\over\partial z}\right)$ (and similarly for $g$): $$ \begin{align} \nabla f(x,y,z)&=\left( {3-x\over(2-x)^2},{3-y\over(2-y)^2},{3-z\over(2-z)^2} \right)\\ \nabla g(x,y,z)&=(1,1,1). \end{align} $$ Setting $\nabla f=\lambda \nabla g$ (to assert that the surfaces' tangent planes are parallel, or their normal lines are parallel, at any local extrema of $f$), we find that $${3-x\over(2-x)^2}={3-y\over(2-y)^2}={3-z\over(2-z)^2}=\lambda.$$ At this point, we would like to be able to conclude $x=y=z$. My argument will use the fact that, although these are quadratics in $x,y,z$, their solution is unique on the domain of interest. If we substitute $t={1\over2-x}$ or $x=2-\frac1t$, we find $$\lambda={3-x\over(2-x)^2}=\frac1{2-x}+\frac1{(2-x)^2}=t+t^2$$ or $$t^2+t-\lambda=0\qquad\text{for}\quad t=\frac1{2-x}.$$ This is a quadratic equation with solution $$t=\frac{-1\pm\sqrt{1+4\lambda}}2.$$ Clearly $x,y,z\in(0,1)$ so $t\in(\frac12,1)$ has unique solution $$t=\frac{-1+\sqrt{1+4\lambda}}2,$$ and therefore $$x=y=z=2-\frac1t=2-\frac{1+\sqrt{1+4\lambda}}{2\lambda}.$$ But now the constraint gives us $x=y=z=\frac13$ (and $t=\frac35,\lambda=\frac{24}{25}$). We can see that $f(\frac13,\frac13,\frac13)=\frac35$ is global minimum for $x,y,z\ge0$ because $f(1,0,0)=1$ is greater.

We can also verify the global minimum by taking second derivatives. Note that $f_{xx}={4-x\over(2-x)^3}>0$ for $x\in(0,2)$ and similarly for $f_{yy},f_{zz}$, while the mixed partials $f_{xy}=f_{yz}=f_{xz}=0$. The Hessian is thus a diagonal matrix with positive entries for $x,y,z\in(0,1)$, and hence is positive definite so that, by the second derivative test, any local extremum will be a global minimum.

bgins
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    A symmetric expression does not mean that the extremal values occur at symmetric inputs. – Arthur Mar 22 '16 at 07:59
  • Yes, @Arthur, of course you're right. But if you know the method of Lagrange multipliers, it seems pretty obvious when the constraint function $g$ has gradient $(1,1,1)$. I added some justification of this. – bgins Mar 22 '16 at 09:24
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Using AM-GM inequality (for $n=6$): Let $f(x,y,z)=\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}$. Take $x_1=\alpha(2-x), x_2=\alpha(2-y), x_3=\alpha(2-z), x_4=\frac{2}{(2-x)}, x_6=\frac{2}{(2-y)}+\frac{2}{(2-z)}$, where $\alpha=\frac{18}{25}$. Then by AM-GM inequality $\frac{1}{6}\left(5\alpha+\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\right)=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\geq \sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}=\sqrt{2\alpha}=\frac{6}{5}$, which gives $\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\geq \frac{36}{5}-5\frac{18}{25}=\frac{18}{5}$. So,

$f(x,y,z)=-3+\frac{2}{(2-x)}+\frac{2}{(2-y)}+\frac{2}{(2-y)}\geq -3+\frac{18}{5}=\frac{3}{5}=-3+6\sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}$, where $x_1=x_2=\ldots=x_6$, which happens when $x=y=z$, since $\alpha\neq 0$, and $\frac{2}{2-x}=\alpha(2-x)$ or $x=\frac{1}{3}$, since $x>0$.

Thus the least value of $f$ is $\frac{3}{5}$.

Singh
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