Finding the minimum value of $y^2+z^2$

Question

If x,y,z are three integers satisfying $x+y+z=13$ and $xy+xz=42$ ,what is the minimum possible value of $y^2+z^2$ ?

How do solve this ? Solution that I thought of : $(x+y+z)/3 >3xyz/(xy+yz+zx)$ but this leaves us with $xyz$ and $yz$ ?

Answer 1

First, you can calculate $x$: $$x(y+z)=42\implies x+{42\over x}= 13\implies...$$

Then use $$y^2+z^2\geq {1\over 2}(y+z)^2$$

Answer 2

$x(y+z)=42$ and $x+(y+z)=13$, which by the Viet's theorem gives $y+z=6$ or $y+z=7$.

For $y+z=6$ we obtain: $$y^2+z^2\geq\frac{1}{2}(y+z)^2=18.$$ The equality occurs for $y=z=3$.

For $y+z=7$ we'll get a greater value.

Answer 3

We can find the value of $x$ as 6 or 7 using the constraints($x+y+z=13$ & $xy+xz=42$). Then we apply Lagrange multipliers method (with given constraints to minimise $y^2+z^2$) and get more info about $y$ and $z$.

We get $y=z$, hence $y=z=3$ when $x=7$ for minimum value of $y^2+z^2$.

Minimum value$=3^2+3^2=18$