<s>[INST] Minimizing expression under constraint I was asked to minimize the expression $x^2+2y^2+3z^2$ under the constraint $xy+yz+zx=1$.
Using the Lagrange-multipliers method, the system to solve, to get eventual extrema, is 
$$ \begin{cases} 2x = \lambda(y+z)  \\ 4y=\lambda(x+z) \\ 6z=\lambda(x+y) \\ xy + yz + xz = 1\end{cases} $$  
I'm having trouble in solving this system. 
As for the minimization by other means, I tried the method suggested by Macavity in this post that uses only AM-GM but failed to find suitable coefficients to get it to work 
Thanks for any advice, suggestion.
 [/INST] We need to find a maximal value of $k$, for which the inequality
$$x^2+2y^2+3z^2\geq k(xy+xz+yz)$$ is true for all reals $x$, $y$ and $z$ or
$$3z^2-k(x+y)z+x^2+2y^2-kxy\geq0,$$ for which we need
$$k^2(x+y)^2-12(x^2+2y^2-kxy)\leq0$$ or
$$(12-k^2)x^2-2(k^2+6k)xy+(24-k^2)y^2\geq0,$$ for which we need $12-k^2>0$ and 
$$(x^2+6k)^2-(12-k^2)(24-k^2)\leq0$$ or
$$k^3+6k^2-24\leq0.$$
Id est, it remains to find a maximal root of the equation
$$k^3+6k^2-24=0.$$
Indeed, let $k=4\cos\alpha-2.$
Thus, we get $$\cos3\alpha=\frac{1}{2},$$ which gives $$\alpha=\pm20^{\circ}+120^{\circ}n,$$ where $n\in\mathbb Z$ and we see that $\alpha=20^{\circ}$ is valid, which gives $k_{max}=4\cos20^{\circ}-2$ and we obtain:
$$\min_{xy+xz+yz=1}(x^2+2y^2+3z^2)=4\cos20^{\circ}-2.$$
</s>