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.

Too long for a comment.   
Since the   inequality is homogeneous, without loss of generality we may suppose $a+b+c=1$  and $x+y+z=1$. Then 
$$(1-x)(1-y)(1-z)=1-x-y-z+xy+xz+yz-xyz=xy+xz+yz-xyz$$
Thus the left hand side of the inequality equals 
$$\frac{1-a}{a(1-x)}+ \frac{1-b}{(1-y)}+ \frac{1-c}{c(1-z)}=$$
$$\frac{1}{a(1-x)}+ \frac{1}{b(1-y)}+ \frac{1}{c(1-z)}-\frac{1}{(1-x)}-\frac{1}{(1-y)}- \frac{1}{(1-z)}=$$
$$\frac{bc(1-y)(1-z)+ac(1-x)(1-z)+ab(1-x)(1-y)}{abc(1-x)(1-y)(1-z)}-
\frac{(1-y)(1-z)+(1-x)(1-z)+(1-x)(1-y)}{(1-x)(1-y)(1-z)}=$$
$$\frac{bc(1-y-z+yz)+ac(1-x-z+xz)+ab(1-x-y+xy)}{abc(xy+xz+yz-xyz)}-
\frac{(1-y-z+yz)+ (1-x-z+xz)+ (1-x-y+xy)}{xy+xz+yz-xyz }=$$
$$\frac{bc(x+yz)+ac(y+xz)+ab(z+xy)}{abc(xy+xz+yz-xyz)}-
\frac{(x+yz)+ (y+xz)+ (z+xy)}{xy+xz+yz-xyz}=$$
$$\frac{bc(x+yz)+ac(y+xz)+ab(z+xy)}{abc(xy+xz+yz-xyz)}-
\frac{1+ xy+yz+xz }{xy+xz+yz-xyz}=$$
$$\frac{\frac 1a(x+yz)+\frac 1b(y+xz)+\frac 1c(z+xy)-1-xy-yz-xz }{xy+xz+yz-xyz}.$$
So we have to show that 
$$\left(\frac 1a(x+yz)+\frac 1b(y+xz)+\frac 1c(z+xy)-1-xy-yz-xz\right)(ax+by+cz)\ge 3(xy+xz+yz-xyz)$$
$$\left(\frac 1a(x+yz)+\frac 1b(y+xz)+\frac 1c(z+xy)\right)(ax+by+cz)\ge 3(xy+xz+yz-xyz)+(1+xy+yz+xz)(ax+by+cz)$$
$$x(x+yz)+ \frac bay(x+yz)+ \frac caz(x+yz)+
\frac abx(y+xz)+ y(y+xz)+ \frac cbz(y+xz)+
\frac acx(z+xy)+ \frac bcy(z+xy)+ z(z+xy)\ge $$
$$3(xy+xz+yz-xyz)+(ax+by+cz)+ax^2y+bxy^2+cxyz+axyz+by^2z+cyz^2+ax^2z+bxyz+cxz^2$$
$$x^2+\frac bay(x+yz)+ \frac caz(x+yz)+
\frac abx(y+xz)+ y^2+ \frac cbz(y+xz)+
\frac acx(z+xy)+ \frac bcy(z+xy)+ z^2+5xyz\ge $$
$$3(xy+xz+yz)+(ax+by+cz)+ax^2y+bxy^2+by^2z+cyz^2+ax^2z+cxz^2$$
or that (because $x^2+y^2+z^2+2xy+2xz+2yz=1$)
$$1+\frac bay(x+yz)+ \frac caz(x+yz)+
\frac abx(y+xz)+ \frac cbz(y+xz)+
\frac acx(z+xy)+ \frac bcy(z+xy)+ 5xyz\ge $$
$$5(xy+xz+yz)+(ax+by+cz)+ax^2y+bxy^2+by^2z+cyz^2+ax^2z+cxz^2$$
Finally, we have to show that 
$$1+\left(\frac ba+\frac ab-5\right)xy+
\left(\frac ca+\frac ac-5\right)xz+
\left(\frac cb+\frac bc-5\right)yz+
\left(\frac ac-a\right)x^2y+\left (\frac bc-b\right)xy^2+
\left (\frac ab-a\right)x^2z+\left (\frac cb-c\right) xz^2+
\left (\frac ba-b\right)y^2z+\left(\frac ca-c\right)yz^2
- (ax+by+cz) + 5xyz \ge 0$$
