$A=\frac{5}{3}$, take $x=y=1$. Equality case:$(x,y)=(1,1)$ $i)$If $x$ is not $2$, we will prove that \[4x^2+3y^2+3 \geq 5xy+5x\]\[LHS=(x^2+3)+3(x^2+y^2)\geq (x^2+3)+6xy \geq 5xy+5x\]\[\iff x^2+3+xy\geq 5x\]It's true because $x^2+xy+3-5x \geq x^2-4x+3=(x-3)(x-1)\geq0$ $ii)$If $x=2$, \[16+3y^2+3\geq 10y+10\]\[3y^2+9\geq 10y\]The last one is true because $3y^2-10y+9=(y-2)(3y-4)+1 \geq0$