What is the Sum of x, y, and z in a Non-Negative Real Number System?

[anemone]

$x,\,y$ and $z$ are non-negative real numbers that satisfy the following system:

$x^2+y^2+xy=3\\y^2+z^2+yz=4\\z^2+x^2+xz=1$

Evaluate $x+y+z$.

 

[DaalChawal]

Is answer 3* ${8 \over 9}^{3/4}$ That 8/9 whole power is 3/4

 

[DaalChawal]

@anemone Can you please tell me the answer or solution?

 

[anemone]

Is answer 3* ${8 \over 9}^{3/4}$ That 8/9 whole power is 3/4

Nope, the correct answer is $\sqrt{7}$.

 

[Opalg]

Spoiler: my solution

$(1)\qquad x^2+y^2+xy=3,\\ (2)\qquad y^2+z^2+yz=4,\\ (3)\qquad z^2+x^2+xz=1.$
Subtract (1) from (2): $z^2 - x^2 + y(z-x) = 1$,
$(z-x)(z+x + y) = 1$,
$s(z-x) = 1$, where $s = x+y+z$. Therefore
$(4)\qquad z = x + \dfrac1s$.
In the same way, subtract (3) from (2): $y^2 - x^2 + z(y-x) = 3$ to get $s(y-x) = 3$ and therefore
$(5)\qquad y = x + \dfrac3s$.
From (4) and (5), $s = x+y+z = 3x + \dfrac4s$ and therefore
$(6)\qquad x = \dfrac13\left(s - \dfrac4s\right)$. Then from (6) and (4),
$(7)\qquad z = \dfrac13\left(s - \dfrac1s\right)$.
Now substitute (6) and (7) into (3): $\dfrac19\left(s - \dfrac4s\right)^2 + \dfrac19\left(s - \dfrac1s\right)^2 + \dfrac19\left(s - \dfrac4s\right)\left(s - \dfrac1s\right) = 1$,
$3s^2 - 15 + \dfrac{21}{s^2} = 9$,
$s^4 - 8s^2 + 7 = 0$,
$\bigl( s^2 - 1\bigr)\bigl( s^2 - 7\bigr) = 0$.
If $s = \pm1$ or $s = -\sqrt7$ then (from (6)) $x$ would be negative. So the only solution for which $x$, $y$ and $z$ are all positive is $s = \sqrt7$.