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

Question

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.

1. 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: $$(ab+bc+ca)^2 \geq 3abc(a+b+c)$$

2. 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?

3. 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.

Answer 1

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

Answer 2

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 $\frac{1}{b(x+z)}+\frac{1}{c(x+y)}\ge\frac{3}{ax+by+cz}$?

Answer 3

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 .