Finding values of $x,y,z$ in three rows of quadratic equations $x^2+xy+y^2=25$
$x^2+xz+z^2=64$
$y^2+yz+z^2=49$ 
$x,y,z>0$  $x+y+z=?$ 
First i have named the rows in order $R_1,R_2,R_3$ as i have tried
$R_3-R_1$   $R_2-R_3$ and  $R_2-R_1$ give reasonable results. Then i had the following
$(z-x).(x+y+z)=24$
$(x-y).(x+y+z)=15$
$(z-y).(x+y+z)=39$
It is obvious that if I add the first two rows it equals to the value of the third row. So i got back to where i started. Maybe the solution can move further from here, i don't know but there is another idea i had.
We know that $\dfrac{a_1+a_2+\cdots+a_n}{n}\ge\sqrt[n]{a_1a_2\cdots a_n}$ In other words Arithmetic mean is greater or equal to Geometric mean.
Proceeding $x^2+xy+y^2\ge\sqrt[3]{x^3y^3}\rightarrow\dfrac{25}{3}\ge xy$ 
Using the same method $\dfrac{64}{3}\ge xz$ and $\dfrac{49}{3}\ge yz$
I multiplied these expressions and came to the conclusion that $\dfrac{280}{9}\ge xyz$ 
$x+y+z\ge 3.\sqrt[3]{xyz}$ Approximately
$x+y+z\ge 11,3313$ That is the approximate value of $\sqrt{129}$ 
In order to fully solve this problem i need to prove that $x+y+z$ is certainly equal. And my question is this;
How can I do that? 
 A: Let $\Delta ABC$ such that $AB=c=7$, $AC=b=8$ and $BC=a=5$.
Easy to see that this triangle is an acute-angled triangle, 
which says that there is a point $F$ inside the triangle for which
$\measuredangle AFB=\measuredangle AFB=\measuredangle AFB=120^{\circ}$. 
This point names Fermat point of the triangle.
Now, let $FA=z$, $FB=y$ and $FC=x$.
Thus, by law of cosines we obtain your system and we need to find a value of $FA+FB+FC$.
Indeed, by law of cosines again we have $\cos\alpha=\frac{8^2+7^2-5^2}{2\cdot8\cdot7}=\frac{11}{14}$.
We'll use the rotation $R^{60^{\circ}}_{A}$ of $\Delta ABC$ (around $A$ on $60^{\circ}$).
Let $R^{60^{\circ}}_{A}(F)=F'$ and $R^{60^{\circ}}_{A}(C)=C'$.
Thus, since $\Delta AFF'$ is an equilateral triangle,
$$x+y+z=CF+BF+AF=C'F'+BF+FF'=C'F'+F'F+FB=C'B,$$
$AF'=AC=b=8$ and $\measuredangle F'AB=60^{\circ}+\alpha$. 
Thus, by law of cosines for $\Delta C'AB$ we obtain: 
$$x+y+z=\sqrt{b^2+c^2-2bc\cos(60^{\circ}+\alpha)}=$$
$$=\sqrt{8^2+7^2-2\cdot8\cdot7\left(\frac{1}{2}\cdot\frac{11}{14}-\frac{\sqrt3}{2}\cdot\frac{5\sqrt3}{14}\right)}=\sqrt{129}.$$
Done!
A: Moving the comments to an answer as a sign of my confidence that the process leads to an answer. This is not a full solution. I'm leaving a bit of the fun for the OP.
Reducing this to elementary symmetric polynomials suggests the following attack. Write $s_1=x+y+z$, $s_2=xy+yz+zx$, $s_3=xyz$, and
$r_1=x^2+xy+y^2$, $r_2=y^2+yz+z^2$, $r_3=z^2+zx+x^2$. Then it is easy to verify that
$$
\begin{aligned}r_1+r_2+r_3&=2s_1^2-3s_2,\\
r_1r_2+r_2r_3+r_3r_1&=s_1^4-3s_1^2s_2+3s_2^2.
\end{aligned}$$
You are given the values of $r_1,r_2,r_3$. And from this pair of equations you should be able to solve $s_1$ by solving $s_2$ from the first and plugging it into the second. Looks like it leads to a biquadratic equation. Hopefully you can eliminate all but one of the alternatives for $s_1$ that pop out.
