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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 $0 ≤ xy + yz + zx − xyz ≤ 2$

4 Answers 4

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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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    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.2012-12-20
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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, therefore we get $3(xyz)^{2/3}, hence $3^3. This contradicts the fact that $x^2+y^2+z^2+xyz=4$

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    If (xyz)^{1/3}, then $xyz\not=0$. Now divide both sides by $(xyz)^{2/3}$. Now we get 3<(xyz)^{1/3}2012-12-20
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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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Let $x=\frac{2a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{2b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ be positives.

Hence, $z=\frac{2c}{\sqrt{(a+c)(b+c)}}$ and the left inequality it's $4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\geq\frac{8abc}{(a+b)(a+c)(b+c)}$ or $\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq2abc,$ which is true because $\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq\sum_{cyc}abc\geq2abc.$ The right inequality.

We need to prove that $4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}-\frac{8abc}{(a+b)(a+c)(b+c)}\leq2$ or $2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq(a+b)(a+c)(b+c)+4abc,$ which is AM-GM: $2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}ab(a+c+b+c)=$ $=\sum_{cyc}(a^2b+a^2c+2abc)=(a+b)(a+c)(b+c)+4abc.$ Done!