Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$

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 $$1+\left(\frac ba+\frac ab-5\right)xy+ \left(\frac ca+\frac ac-5\right)xz+ \left(\frac cb+\frac bc-5\right)yz+ \left(\frac ac-a\right)x^2y+\left (\frac bc-b\right)xy^2+ \left (\frac ab-a\right)x^2z+\left (\frac cb-c\right) xz^2+ \left (\frac ba-b\right)y^2z+\left(\frac ca-c\right)yz^2 - (ax+by+cz) + 5xyz \ge 0$$