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Let $x$, $y$, $z$ be three positive real numbers satisfying \begin{equation} x + y +z + 1 =4xyz.\tag{1} \end{equation} Prove that \begin{equation} xy + yz + zx \geqslant x + y + z.\tag{2} \end{equation} I don't know how to start?

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up vote 3 down vote accepted

Here is a boring, uninspiring but elementary proof ^_^. By $(1)$ and A.M.$\ge$G.M., we get $xyz = (x+y+z+1)/4 \ge (xyz)^{1/4}$. Hence $xyz\ge1$. WLOG, suppose $0<x\le y\le z$. There are three cases. Case 1: $x\ge1$. Clearly $(2)$ holds. Case 2: $0<x<1\le y< z$. Then $(2)$ also holds because $$1-xyz\le0\le(1-x)(1-y)(1-z) = 1 - (x+y+z) + (xy+yz+zx) - xyz.$$ Case 3: $0<x\le y<1<z$. Then \begin{align*} (2)&\Leftrightarrow z(x+y) + xy \ge x+y+z,\\ &\Leftrightarrow z(x+y-1) + xy - x-y\ge0,\\ &\Leftrightarrow z(x+y-1)-1 + (x-1)(y-1) \ge0. \end{align*} So it suffices to show that $z(x+y-1)-1\ge0$. Since $x,y,z$ are positive, we must have $4xy>1$, otherwise the LHS of $(1)$ will be strictly greater than the RHS. Now $(1)$ implies that $z=(x+y+1)/(4xy-1)$. Therefore $$ z(x+y-1)-1 =\frac{(x+y+1)(x+y-1)-(4xy-1)}{4xy-1} =\frac{(x-y)^2}{4xy-1} \ge0 $$ and we are done. From the details of above three cases, it can be shown that equality in $(2)$ holds iff $x=y=z=1$.

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Firstly, I'm going to relabel $x$,$y$ and $z$ to $\alpha$,$\beta$ and $\gamma$. We then consider the cubic polynomial $$ f(x) = ax^3 + bx^2 + cx + d$$ Supposing this has roots $\alpha$,$\beta$ and $\gamma$, by using the factor theorem, expanding the brackets and collecting together coeffecients we get the relationships: $$ -\frac{d}{a} = \alpha\beta\gamma $$ $$ \frac{c}{a} = \alpha\beta + \beta\gamma + \gamma\alpha$$ $$ - \frac{b}{a} = \alpha + \beta + \gamma $$

We then have a condition on the coeffecients of the polynomial to consider instead. The inequality then should hopefully come out when considering suffecient conditions for a cubic polynomial to have three real positive roots.

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Very elegant trick. – Rhymoid Jan 8 '13 at 15:13

Let $F(x,y,z)=x+y+z+1-4xyz$ and $G(x,y,z)=xy+yz+zx-x-y-z$. Then we wish to minimize $G$ subject to $F=0$, and so can use lagrange multipliers. Then the following are zero at a critical point: $$F_xG_y-F_yG_x=(x-y)(2z-1)^2,$$ $$F_yG_z-F_zG_y=(y-z)(2x-1)^2,$$ $$F_zG_x-F_xG_z=(z-x)(2y-1)^2.$$ Here we cannot have $x=y=z=1/2$ and satisfy $F=0$, so at any critical point two of the variables are equal. By symmetry put $x=y=s$ and $z=t$; then $G=0$ gives $t=1/(2s-1)$. And in this case $F$ becomes $s^2-2s+1=(s-1)^2 \ge 0.$

Note that this technique applies to the case $x,y,z \ge 0$, since $F=0$ is inconsistent with any of $x,y,z$ being $0$. So the min of $G$ must occur at a critical point, and that min is zero as above, making the desired inequality true.

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Don't we need to discuss $\lambda=0$ and rule it out at the beginning? – Maesumi Jan 8 '13 at 15:38
@Maesumi : The gradient of $G$ is $(y+z-1,x+z-1,x+y-1)$. So if it’s zero, all the coordinates $x,y,z$ are equal to $\frac{1}{2}$, which is impossible. Similarly, the gradient of $F$ is $(1-4yz,1-4xz,1-4xy)$. So if it's zero, all the coordinates $x,y,z$ are still equal to $\frac{1}{2}$. This shows that none of the gradients is zero. – Ewan Delanoy Jan 8 '13 at 15:44
I think the two gradient vectors are proportional at a critical point, and this is equivalent to the three equations displayed. – coffeemath Jan 8 '13 at 15:46
@cofeemath : actually, we might have an absolute minimum for $G$, independently of $F$, in that case the gradient of $G$ would be zero (this is a separate case to check every time). My comment above treats this. – Ewan Delanoy Jan 8 '13 at 15:48
If the gradient of $G$ is the zero vector $(0,0,0)$, then the three equations displayed in my answer involving the partials all hold (each being a difference of products of partials, with each product including one of the partials of $G$). So the quantities to the right of those equations are all zero, leading to at least two of $x,y,z$ being equal at a critical point. That's all I used about the critical point in the argument. – coffeemath Jan 9 '13 at 14:47

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