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.

Partial Proof:
If we start with your substitution at 3) and remark that we have with your condition :
$$ux+yv+zw=(x+y+z)(u+v+w)-u(y+z)-v(z+x)-w(x+y)=(x+y+z)-u(y+z)-v(z+x)-w(x+y)$$
So if we put :
$p=y+x$$\quad$$i=w(x+y)$
$q=z+x$$\quad$$k=v(z+x)$
$r=y+z$$\quad$$j=u(y+z)$
We get :
$$\frac{1}{i}+\frac{1}{j}+\frac{1}{k}+\frac{-3}{\frac{p+r+q}{2}-i-j-k}\geq \frac{1}{p}+\frac{1}{q}+\frac{1}{r} $$
With the condition :
$$\frac{i}{p}+\frac{k}{q}+\frac{j}{r}=1. $$
Now the idea is to use the relation between the incircle of center I and the side of a triangle ABC :
$$\frac{IA^2}{CA.AB}+\frac{IB^2}{BC.AB}+\frac{IC^2}{CA.BC}=1$$
So we put :
$p=CA.AB$$\quad$$i=IA^2$
$q=BC.AB$$\quad$$k=IB^2$
$r=CA.BC$$\quad$$j=IC^2$
We get :
$$\frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2}+\frac{-3}{\frac{CA.AB+BC.AB+CA.BC}{2}-IA^2-IB^2-IC^2}\geq \frac{1}{CA.AB}+\frac{1}{BC.AB}+\frac{1}{CA.BC} $$
Furthermore we have the following relation :
$$\frac{1}{IA^2}+\frac{1}{IB^2}+\frac{1}{IC^2}=\frac{1}{r^2}-\frac{1}{2rR}$$
$$IA^2+IB^2+IC^2=s^2+r^2+8rR$$
$$CA.AB+BC.AB+CA.BC=s^2+(4R+r)r$$
$$\frac{1}{CA.AB}+\frac{1}{BC.AB}+\frac{1}{CA.BC}=\frac{1}{2rR} $$
Where $s$ denotes the semi-perimeter , $r$  the radius of the incircle and R the radius of the excircle 
Finally we have :
$$\frac{1}{r^2}-\frac{1}{2rR}+\frac{-3}{-(s^2+r^2+8rR)+0.5(s^2+(4R+r)r)}\geq \frac{1}{2rR} $$
I finally found a theorem of Blundon wich characterizes the existence of a scalene triangle with a necessary and sufficient condition on $r$,$R$,$s$ .See here.But I don't know how to use it to have a condition on the inequality . Maybe somebody could do that .
