Maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$ under given conditions Consider $a,b,c \in \mathbb{R}-\{0\}$. A real valued function is defined as $$f(x,y,z)=2^y3^za^xb^yc^z+2^z3^xa^yb^zc^x+2^x3^ya^zb^xc^y$$ where $x,y,z \in \mathbb{Z}$ and $f(1,0,0)=4$ and $f(2,0,0)=6$. Find the maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$.
Using given equations I got
$a+2b+3c=4$ and $a^2+4b^2+9c^2=6$ and also $2ab+6bc+3ac=5$. I have also obtained $a^2+b^2+c^2 \geq \displaystyle\frac{8}{7}$ but how to proceed further? Could someone help me with this?

Sorry, just a partial answer:
If you substitute $x=a$, $y=2b$, $z=3c$, then your first two equations are $x+y+z=4$ and $x^2+y^2+z^2=6$.  These surfaces intersect in a small circle in the first octant, So that shows $a, b, c$ are all positive.  Your last inequality shows that at most one of them is greater than $1$ so the final answer is either $0$ or $1$.   You can work out that the center of the circle is at $(4/3,4/3,4/3)$ in $xyz$-space, and has radius $\sqrt{2/3}.$
I think that if you play with that (tedious) geometry a bit, you can prove that $a,b$, and $c$ are all less than one.  (I graphed the 3 surfaces in Maple, and this is certainly the case.)
