<s>[INST] How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq \frac{21}{4}$$
 [/INST] 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.
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