Minimizing expression under constraint

Question

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.

Answer 1

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: $$\min_{xy+xz+yz=1}(x^2+2y^2+3z^2)=4\cos20^{\circ}-2.$$

Answer 2

Using Maple's Groebner basis package, the system $$ \begin{cases} w=x^2+2y^2+3z^2\\[4pt] 2x = \lambda(y+z)\\[4pt] 4y=\lambda(z+x)\\[4pt] 6z=\lambda(x+y)\\[4pt] xy+yz+zx=1\\ \end{cases} $$
of $5$ equations in $5$ unknowns yields the equation $$w^3+6w^2-24=0$$ which has $3$ real roots, but only one positive real root $$w\approx 1.758770483$$ which is the required minimum value.

The minimum is realized at the two points $$(x_1,y_1,z_1)$$ $$(-x_1,-y_1,-z_1)$$ where $x_1,y_1,z_1$ are the unique positive real roots of the respective equations $$9x^6+9x^4-9x^2-1=0$$ $$9y^6+9y^4-1=0$$ $$81z^6+81z^4-9z^2-1=0$$ yielding $$(x_1,y_1,z_1)\approx (.8294933740, .5414133484, .4018517131)$$