Minimum

Relevant wiki: Lagrange Multipliers

Here is a calculus method,

By Lagrange Multipliers method we set $g(x,y,z)-c=x+2y+3z-7$

$f(x,y,z)=x^2+y^2+z^2$ & $\mathfrak{L}(x,y,z,\lambda)=f(x,y,z)+\lambda(g(x,y,z)-c)$

Now we have $\Delta_{x,y,z,\lambda}\mathfrak{L}(x,y,z,\lambda) = (\frac{\partial{L}}{\partial{x}},\frac{\partial{L}}{\partial{y}},\frac{\partial{L}}{\partial{z}},\frac{\partial{L}}{\partial{\lambda}})$

$\Delta_{x,y,z,\lambda}\mathfrak{L}(x,y,z,\lambda) = (2x+\lambda,2y+2\lambda,2z+3\lambda,x+2y+3z)$

$\Delta_{x,y,z,\lambda}\mathfrak{L}(x,y,z,\lambda) = 0\implies \begin{cases} 2x+\lambda=0 \\ 2y+2\lambda=0 \\ 2z+3\lambda \\ x+2y+3z=7 \end{cases}$

Solving we get the critical point $(x,y,z)=(\frac{1}{2},1,\frac{3}{2})$

Moreover This does not assure us it's a point of minimum, but just a critical point. We see $f(\frac{1}{2},1,\frac{3}{2})=3.5$ .

Taking the Hessian Matrix defined by ,

$\begin{aligned} \Delta^2f(x,y,z) = \begin{pmatrix} 2&0&0\\0&2&0\\0&0&2\end{pmatrix}\end{aligned}$ , we readily observe it's positive definite. So $f(x,y,z)$ has a local minimum & it's indeed the only stationary point we have. Hence $f(x,y,z)\ge f(\frac{1}{2},1,\frac{3}{2}) = 3.5$

Using Cauchy-Schwarz inequality, we have:

$\begin{aligned} (1^2+2^2+3^2)(x^2+y^2+z^2) & \ge (x+2y+3z)^2 \\ (1+4+9)(x^2+y^2+z^2) & \ge 7^2 \\ \implies x^2+y^2+z^2 & \ge \frac {49}{14} = \frac 72 = \boxed{3.5} \end{aligned}$

Apply the Cauchy-Schwarz inequality.
(/(1^2+2^2+3^2/))*(/(x^2+y^2+z^2/))≥/(49/)
/(x^2+y^2+z^2/)≥/(3.5/)
therefore, the minimum is 3.5