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?

$a+2b+3c=4$ ...(i)
$a^2+4b^2+9c^2=6$ ...(ii)
Given we have to find the maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$ which has equal of $a^2, b^2, c^2$, we should maximize $a^2$ in (ii).
The minimum value of $a^2, b^2, c^2$ can be $0$ each. So individually, max value of
$a^2 = 6, b^2 = \dfrac{3}{2}, c^2 = \dfrac{6}{9}$ ...(iii)
So $\lfloor c \rfloor^2$ cannot be any value other than $0$.
Now, let's maximize $\lfloor a \rfloor^2$. From (iii), max value of $a = \sqrt6$ but any value greater than $2$ (and less than $3$) does not help. So, let's go with $a^2 = 4$.
Now, $4b^2+9c^2 = 2$. Even with $c^2$ as $0$, $b \lt 1$.
So, maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2 = 4$
