Find $x,y,z\in\mathbb N$ where $x,y,z$ are lengths of a triangle and $x\le y\le z$, $x^2+y^2+z^2+126\le13(x+y+z)$, and $x^3+y^2+z=272$.

Question

Question: Find $x,y,z\in\mathbb N$ where $x,y,z$ are lengths of a triangle and $x\le y\le z$, $x^2+y^2+z^2+126\le13(x+y+z)$, and $x^3+y^2+z=272$.

What I tried so far without any luck: \begin{align} &272=x^3+y^2+z> x+y+z>2z\rightarrow z<141\\ &272=x^3+y^2+z> x+y+z\ge3x\rightarrow x\le90\\ &x+y+z<x^2+y^2+z^2\le13(x+y+z)-126\rightarrow x+y+z>\frac{126}{12}\\ &x^2+y^2+z^2+126\le13(x+y+z)<13(x^2+y^2+z^2)\rightarrow x^2+y^2+z^2>\frac{126}{12}\\ &272=x^3+y^2+z>x^3\rightarrow x\le6\\ &272=x^3+y^2+z>y^2+y\rightarrow y\le15\\ &z<x+y\le21\\ &\cdots\\ \end{align}

Answer 1

The first condition gives $$(2x-13)^2+(2y-13)^2+(2z-13)^2\leq3,$$ which gives $$\{2x-13,2y-13,2z-13\}\subset\{1,-1\}$$ and the rest is smooth.