Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$ 
Suppose that $a,b,c,x,y,z$ are all positive real numbers. Show that
  $$\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$$

Below are what I've done, which may be misleading.


*

*I've tried to analyze when the equality holds：


1.1 Under the condition that $a=b=c$, it reduces to
$$\frac{2}{y+z}+\frac{2}{z+x}+\frac{2}{x+y}\geq \frac{9}{x+y+z}$$
1.2 Under the condition that $x=y=z$, it reduces to
$$\frac{b+c}{2a}+\frac{c+a}{2b}+\frac{a+b}{2c}\geq 3$$
Both are easy to verify. However, $ax+by+cz$ is not easy to deal with. Is there any famous inequality that I can use here? 
1.3 Under the condition that $(x,y,z)$ and $(a,b,c)$ are in proportion, i.e. $x=at, y=bt, z=ct$ for some $t>0$, the inequality reduces to
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3(a+b+c)}{a^2+b^2+c^2}$$
which can be proved by using Newton's inequality:
$$(ab+bc+ca)^2 \geq 3abc(a+b+c)$$


*I’ve also tried to construct a function
$$f(x,y,z)=\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}- \frac{3(a+b+c)}{ax+by+cz}$$ 
and analyze its global minimum. But the first-order condition is complicated
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{c+a}{ab(z+x)^2}+\frac{a+b}{ca(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ab(y+z)^2}+\frac{a+b}{bc(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ca(y+z)^2}+\frac{c+a}{bc(z+x)^2}$$
Maybe some convexity can be used here?

*Substitution has been considered. Let
$$u=\frac{a}{a+b+c},v=\frac{b}{a+b+c}, w=\frac{c}{a+b+c}$$
Then $u,v,w>0$ and $u+v+w=1$, which are just weights we assign to $x,y,z$. The inequality becomes
$$\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}\geq \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}$$
Define another function in variables $u,v,w$
$$g(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
and consider the constrained optimization problem:
$$\min g(u,v,w)\\ \text{s.t. } u>0\\ v>0 \\w>0\\ u+v+w=1$$
The corresponding Lagrangian function can be
$$L_{\lambda}(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}+\lambda(u+v+w-1)$$
And the first-order conditions give
$$\frac{1}{u^2(y+z)}-\frac{3x}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{u(y+z)}-\frac{3ux}{(ux+vy+wz)^2}=\lambda u\\ \frac{1}{v^2(z+x)}-\frac{3y}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{v(z+x)}-\frac{3vy}{(ux+vy+wz)^2}=\lambda v\\ \frac{1}{w^2(x+y)}-\frac{3z}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{w(x+y)}-\frac{3wz}{(ux+vy+wz)^2}=\lambda w\\ u+v+w=1$$
Summing up yields
$$\lambda=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
None of the methods I tried seems to work, and now I even doubt the truth of the inequality.

This is not an answer, but putting code in a comment doesn't work well either.
Regarding your doubts about the truth of the inequality, I am still inclined to think it holds true. You can try to look for counterexamples with this code. In case python thinks the inequality doesn't hold, it outputs $a,b,c,x,y,z$ and the difference between the two sides. So far I have only found differences of $10^{-15}$ and smaller with it, which is just python lacking sufficient precision.
sta = 3.14
sto = 4.14
ste = .1
def frange(start, stop, step):
    i = start
    while i < stop:
        yield i
        i += step
for a in frange(sta,sto,ste):
    for b in frange(sta,sto,ste):
        for c in frange(sta,sto,ste):
            for x in frange(sta,sto,ste):
                for y in frange(sta,sto,ste):
                    for z in frange(sta,sto,ste):
                        if (1.*(b+c)/(a*(y+z))+1.*(c+a)/(b*(z+x))+1.*(a+b)/(c*(x+y))<3.*(a+b+c)/(a*x+b*y+c*z)):
                            print a,b,c,x,y,z,1.*(b+c)/(a*(y+z))+1.*(c+a)/(b*(z+x))+1.*(a+b)/(c*(x+y))-3.*(a+b+c)/(a*x+b*y+c*z)

Also, since the inequality can be reshaped to
$$\frac{a}{b(x+z)}+\frac{b}{a(y+z)}+\frac{a}{c(x+y)}+\frac{c}{a(y+z)}+\frac{b}{c(x+y)}+\frac{c}{b(x+z)}\ge\frac{3a}{ax+by+cz}+\frac{3b}{ax+by+cz}+\frac{3c}{ax+by+cz},$$
Maybe we can show that $\frac{1}{b(x+z)}+\frac{1}{c(x+y)}\ge\frac{3}{ax+by+cz}$?
