<s>[INST] Proving infinity limit of a multivariable function I have a function $f(x,y,z): \mathbb{R}^3\rightarrow \mathbb{R}$ which is $f(x,y,z)=3x^2+z^2+y^2-2xy+14$.
I'm trying to show that $f(x,y,z)\rightarrow \infty$ when $||(x,y,z)|| \rightarrow \infty$.
(Or formally: $\forall_{M>0}\exists_{R>0}\forall_{(x,y,z) \in \mathbb{R}^3}:||(x,y,z)||>R \rightarrow f(x,y,z)>M $)
So in the process of building the proof I'm trying to find a suitable $R$, and for that I'm trying to express $f(x,y,z)$ in terms of $||(x,y,z)||$ or $||(x,y,z)||^2$.
Try 1: 
Let $(x,y,z)$ be such that $||(x,y,z)||>R$, then $f(x,y,z) = 3x^2+z^2+y^2-2xy+14$
$= x^2+z^2+y^2 +2x^2-2xy+14 \geq R^2+2(x^2-xy)+14 $
$\geq R^2-2(\sqrt{2x^2} \sqrt{x^2+y^2})+14 $
$\geq R^2-2(\sqrt{2x^2} \sqrt{x^2+y^2+z^2})+14 $
$\geq R^2-2(\sqrt{2x^2} R)+14 = R^2-2\sqrt{2}|x|R+14$.
From which I can't continue since if I try to bound $x$ by $R$ I get a negative expression.
Try 2: 
Using $xy \leq \frac{x^2+y^2}{2}$, we get that 
$f(x,y,z) = 3x^2+z^2+y^2-2xy+14 \geq 2x^2+z^2+14$, 
from which I can't proceed since I can't express $R$ without $y$.
Try 3: 
Let $R=\sqrt{3}M$, and since $||(x,y,z)||>R$ then $x^2+y^2+x^2>R^2=3M^2$. 
Meaning that at least one of $x^2$, $y^2$, $z^2$ has to be greater or equal $M^2$.
$f(x,y,z) = 3x^2+z^2+y^2-2xy+14 = 2x^2+z^2+(y-x)^2+14$. 
If $x^2>M^2$ or $z^2>M^2$ then obviously $f(x,y,z)>M$ and we're done. But I don't know how to handle the case of $y^2>M^2$.
Any help would be appreciated!
 [/INST] Solved! Given $M>0$, setting $R=\sqrt{2M}$, then for all vectors $(x,y,z) \in \mathbb{R}^3$ for which $||(x,y,z)||>R$, it holds that:
$f(x,y,z)=3x2+z2+y2−2xy+14=$
$0.5(x^2+y^2+z^2)+0.5(x^2+z^2+(2x-y)^2+28)$
$\geq 0.5(x^2+y^2+z^2) > M$
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