Minimize this real function on $\mathbb{R}^{2}$ without calculus? When it comes to minimizing a differentiable real function, calculus comes into play immediately. If $f: (x,y) \mapsto (x+y-1)^{2} + (x+2y-3)^{2} + (x+3y-6)^{2}$ on $\mathbb{R}^{2}$, and if one is asked to find the minimum of $f$ along with the minimizer(s), is it possible to do that without calculus? The three equations do not admit a common solution; besides, I was not seeing an elementary inequality that might be useful at this point. Although this question itself may not be very interesting, I am interested in knowing an elegant way for the (more or less recreational) minimization.
 A: \begin{align*}
f(x,y)&=3x^2+12xy+14y^2-20x-50y+46\\
&=3(x+2y)^2+2y^2-20(x+2y)-10y+46\\
&=\frac13(3x+6y-10)^2+2y^2-10y+\frac{38}3\\
&=\frac13(3x+6y-10)^2+\frac12(2y-5)^2+\frac16
\end{align*}
The minimum value is $\dfrac16$. It happens when $\displaystyle (x,y)=\left(-\dfrac53,\dfrac52\right)$.
A: In general, any quadratic function $\ f\ $ on $\ \mathbb{R}^n\ $ can be written as
$$
f\left(x\right) = x^\top A x + b^\top x + c\ ,
$$
where $\ A\ $ is a symetric $\ n\times n\ $ matrix, $\ b\ $ an $\ n\times 1\ $ column vector and $\ c\ $ a constant.  A minimum exists if and only if $\ A\ $ is positive definite or semidefinite and $\ b\ $ lies in its column space.  If these conditions are satisfied, and $\ b=-2 Ax_0\ $, then
$$
f\left(x\right) = (x-x_0)^\top A\, (x-x_0) + c-x_0^\top A x_0\ ,
$$
and has a minimum value $\ c-x_0^\top A x_0\ $ when $\ x=x_0\ $.
For the function $\ f\ $ given in the question,
$$
f\left(x,y\right) = \pmatrix{x&y}^\top\pmatrix{3&6\\6&14}\pmatrix{x\\y} + \pmatrix{-20&-50}\pmatrix{x\\y}+46\ ,
$$
and we have
$$
\pmatrix{-20\\-50} = -2\pmatrix{3&6\\6&14}\pmatrix{-\frac{5}{3}\\ \frac{5}{2}}\ ,
$$
leading to the same result as given in the other answers.
A: By C-S
$$f(x,y)=\frac{1}{6}(1+4+1)\left((1-x-y)^2+\left(x+2y-3\right)^2+(6-x-3y)^2\right)\geq$$
$$=\frac{1}{6}\left(1-x-y+2x+4y-6+6-x-3y\right)^2=\frac{1}{6}.$$
The equality occurs for
$$(1,2,1)||(1-x-y,x+2y-3,6-x-3y),$$ id est, for
$$(x,y)=\left(-\frac{5}{3},\frac{5}{2}\right),$$ which says that $\frac{1}{6}$ is a minimal value.
A: See How to Find the Vertex of a Quadratic Equation.
$\tag 1 f(x,y) = 3 x^2 + 4 x (3 y - 5) + 2 (7 y^2 - 25 y + 23)$
Let
$$\tag 2 x = \frac{-4(3y-5)}{6}$$
(Vertex = $\frac{-b}{2a}$).
and plug back into $\text{(1)}$, giving
$M(y) = 1/2 (2 y - 5)^2 + 1/6$
as the quantity to be minimized.
So at $y = \frac{5}{2}$ the minimum of $\frac{1}{6}$ is achieved.
Plugging $\frac{5}{2}$ into $\text{(2)}$ (certainly easier than using $\text{(1)}$ again), we get
$$\tag 3 x = \frac{-4(3(\frac{5}{2})-5)}{6} = -\frac{5}{3}$$
So
$$ (x,y) = (-\frac{5}{3},\frac{5}{2})$$
A: Here is a geometric answer. This is slightly cheating since the duality between planes and normals is essentially what one obtains from the optimality conditions from calculus.
Note that $n=(1,-2,1)^T$ is orthogonal to the plane spanning $(1,1,1)^T, (1,2,3)^T$ and
we are trying to find the closest point to $b=(1,3,6)^T$. From the closest point we can find $x,y$.
The plane is defined by $\{ x | n^T x =0 \}$. Let $p$ denote the closest point.
We must have $b-p=tn$ for some $t$.
Since $b-p$ is orthogonal to the plane, we have $n^Tp = 0$, or $t = {n^Tb \over n^T n} = {1 \over 6}$ and so
$p={1 \over 6}(5,20,35)^T$.
Now we can solve for $x,y$ to get $(x,y)^T = {1 \over 6}(-10,15)^T$.
A: Here's my solution without calculus (not sure how elegant it is though). 
We make a few changes of variable; firstly replace $x$ with $x + 3$, and then let $a = x+2y, b = y$. We obtain $(a-b-2)^2 + a^2 + (a+b+3)^2$, and maximising this over $a$ and $b$ allows us to recover $x$ and $y$. 
Note that we have a $(a-b-2)^2$ term and a $(a+b+3)^2$ term; one has $b$ and one has $-b$ so the sum is maximised when they are closest together, i.e. $b = -\frac{5}{2}$ both squares become $(a+ \frac{1}{2})^2$. So we now need to minimise $2(a+ \frac{1}{2})^2  + a^2 = 3a^2 + a + \frac{1}{2}$, but since this is a quadratic this minimum occurs at $a = \frac{-1}{6}$, and so we simply substitute back to find $x, y$.
A: Let 
$$3\,x^{\,2}+ 12\,xy+ 14\,y^{\,2}- 20\,x- 56\,y+ 46- \frac{1}{6}= \frac{1}{3}(\,3\,x+ 5\,)(\,3\,x+ 12\,y- 25\,)+  \frac{7}{2}(\,5- 2\,y\,)^{\,2}$$
$$18(3 x^{ 2}+ 12 xy+ 14 y^{ 2}- 20 x- 56 y+ 46- \frac{1}{6})= 7(3 x+ 6 y- 10)^{ 2}- (3 x+ 5)(3 x+ 12 y- 25)$$
$$\therefore\,3\,x^{\,2}+ 12\,xy+ 14\,y^{\,2}- 20\,x- 56\,y+ 46- \frac{1}{6}\geqq 0$$
Furthermore
$$\because\,{\rm discriminant}[\,3\,x^{\,2}+ 12\,xy+ 14\,y^{\,2}- 20\,x- 56\,y+ 46- \frac{1}{6},\,x\,]= -\,6(\,5- 2\,y\,)^{\,2}\leqq 0$$
