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Given that $x, y, z$ are nonnegative real numbers such that :

$$x^2 + y^2 + z^2 + xyz = 4$$

Prove that $0 ≤ xy + yz + zx − xyz ≤ 2$

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3 Answers 3

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Here's a straight-forward way, which is not very elegant, but is on the other hand very general, and does not require problem-specific tricks.

We want to calculate bounds for the function $$f=x y + y z+ z x - x y z\ ,$$ under the constraint $$g=x^2 + y^2 + z^2 + x y z-4=0\ .$$

For this, we introduce the Lagrange multiplier $\lambda$, and look for points for which $$\nabla(f-(\lambda-1) g)=\left( \begin{array}{c} y+z-2 x (\lambda -1)-y z \lambda \\ x+z-2 y (\lambda -1)-x z \lambda \\ x+y-2 z (\lambda -1)-x y \lambda \\ \end{array} \right)=0$$ The choice $\lambda-1$ rather than $\lambda$ is arbitrary and will be convenient in the following. Solving the above equation for $x,y,z$ requires some work, and the result gives 2 families of solutions:

  1. $x=y=z=\frac{4}{\lambda }-2$
  2. $x =y=2-\frac{1}{\lambda } ,\ z= \frac{3-8 \lambda +4 \lambda ^2}{2 \lambda -2 \lambda ^2}$ (and cyclic permutations of this).

Solving for $\lambda$, the first solution satisfies $g$ only when $\lambda=\frac{3}{4}$, and then $x=y=z=1$ and $f=2$. Doing the same trick for the second one, one gets only negative results for $x,y$ or $z$, so we can forget about that (for example, the point $x=y=\frac{2}{3},\ z=-2$ satisfies $g$ but not the non-negativity constraint).

We see that for $x,y,z>0$, there is only one critical point of $f$ under the constraint $g$, at the point $x=y=z=1$. A simple check shows that this is a global maximum, since the point $x=y=0, z=2$ satisfies $g$ and has $f=0<2$. You now just have to check the boundaries, i.e. WLOG $x=0$, but this gives immediately $f=yz$ which is non-negative by assumption. therefore $0\le f\le 2$.

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Don`t you need to justify the use of lagrangian mulitplicators? I mean that your constraint does not define a closed smooth submanifold. –  wspin Dec 20 '12 at 14:19
    
I did, in way. There are actually four constraints: $x>0, \ y>0, \ z>0$, and $g$. Each of them is smooth, and for the first three the check is easy. –  yohBS Dec 20 '12 at 14:25
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For the left part of the inequality:

By the AGM-inequality, we have: $$3(xyz)^{2/3}=3((xy)(xz)(yz))^{1/3}\leq xy+xz+yz$$ Suppose that $xy+xz+yx<xyz$, therefore we get $3(xyz)^{2/3}<xyz$, hence $3^3<xyz$. This contradicts the fact that $x^2+y^2+z^2+xyz=4$

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where did $3^3$ come from? –  John Chang Dec 20 '12 at 22:31
    
If $(xyz)^{1/3}<xyz$, then $xyz\not=0$. Now divide both sides by $(xyz)^{2/3}$. Now we get $3<(xyz)^{1/3}$ –  Amr Dec 20 '12 at 22:38
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use this $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$ and $$x^2+y^2+z^2+xyz=4$$ then we set $$x=2\cos{A},y=2\cos{B},z=2\cos{C}$$ $$\Longleftrightarrow 0\le 4(\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{A}\cos{C})-8\cos{A}\cos{B}\cos{C}\le 2$$ and since $$\cos{A}\cos{B}\cos{C}=\dfrac{s^2-(2R+r)^2}{4R^2},\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{C}\cos{A}=\dfrac{s^2+r^2-4R^2}{4R^2}$$

then $$\Longleftrightarrow 0\le\dfrac{s^2+r^2-4R^2}{R^2}-\dfrac{2}{R^2}\left(s^2-(2R+r)^2\right)\le 2$$ and left hand it suffices that $$s^2+r^2-4R^2-2s^2+2(2R+r)^2\ge 0$$ $$\Longleftrightarrow 3r^2+8Rr+4R^2\ge s^2$$

use the Gerrseten inequality: $$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$ then $$\Longleftrightarrow 4R^2+4Rr+3r^2\le 3r^2+4R^2+8Rr$$ $$\Longleftrightarrow 4Rr\ge 0$$is obvious. and the Right Hand, $$\Longleftrightarrow s^2+r^2-4R^2-2(s^2-(2R+r)^2)\le 2R^2 $$ $$\Longleftrightarrow 2R^2+8Rr+3r^2\le s^2$$ since $$s^2\ge r(16R-5r)$$, it follow that $$r(16R-5r)\ge 2R^2+8Rr+3r^2$$ $$\Longleftrightarrow (R-2r)^2\ge 0$$ is obivous.

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