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.