For the second and the third question,a generalized version is presented here:
Let $a,b,c>0$ and $a+b+c=3$,then
$f(a,b,c)=(\sum\frac{ab}{c})+\lambda abc\geq3+\lambda$
for $1\leq \lambda\leq9/4$,where the constant $9/4$ is optimal.
It is easy to see that 9/4 is optimal(a simple comparation between $f(1,1,1)$ and $f(2,1/2,1/2)$).
Proof:
Without loss of generality,we assume that $a\geq b\geq c$.Then $1\leq a<3$.
$f(a,b,c)=a(\frac{b}{c}+\frac{c}{b})+\frac{bc}{a}+\lambda abc=\frac{a(b+c)^2}{bc}+(\lambda a+\frac{1}{a})bc-2a$
Recall that $a+b+c=3$,then
$f(a,b,c)=\frac{a(3-a)^2}{bc}+(\lambda a+\frac{1}{a})bc-2a$
Given $p,q>0$,it is obvious that function $u(x)=px+q/x$ is monotone decreasing on interval $(0,\sqrt{q/p}]$.We take $(\lambda a+\frac{1}{a},a(3-a)^2)$ as $(p,q)$,then $f(a,b,c)$ is a monotone decreasing function(for $bc$) on the interval $(0,\sqrt{q/p}]$ with $a$ fixed.
AM-GM inequality suggests that $bc\leq\frac{(b+c)^2}{4}=\frac{(3-a)^2}{4},$ and it is natural to test whether $(b+c)^2/4\leq\sqrt{q/p}$,i.e., $\frac{(3-a)^2}{4}\leq\frac{a(3-a)}{\sqrt{\lambda a^2+1}}$
It suffices to show that $\frac{(3-a)^2}{16}\leq\frac{a^2}{\lambda a^2+1},$i.e.,
$\lambda\frac{(3-a)^2}{16}+\frac{1}{\lambda a^2+1}\leq 1$.
Because $1\leq a<3$,$LHS\leq\lambda/4+1/(\lambda+1)\leq 1$ for every $1\leq \lambda\leq 3$
Hence $f(a,b,c)$ achieve its minimum(for fixed $a$) when $bc=(3-a)^2/4$.
$\min_{a fixed}f(a,b,c)=(\lambda a+\frac{1}{a})\frac{(3-a)^2}{4}+2a$.
We just need to show that
$(\lambda a+\frac{1}{a})\frac{(3-a)^2}{4}+2a\geq 3+\lambda$
for every $1\leq \lambda\leq 9/4$,which is equivalent to show that
$\lambda(a-1)^2(a^2-4a+\frac{9}{\lambda})\geq 0$
for every $1\leq \lambda\leq 9/4$.
It suffices to prove that $(a^2-4a+\frac{9}{\lambda})\geq 0$ for for every $1\leq \lambda\leq 9/4$,and it is quite obvious.
Q.E.D.