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

  1. 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?

  2. 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.

  • $\begingroup$ The inequality looks doubtful but another inequality looking similar can be proved as follows: to simplify things a bit we may suppose a+b+c=1. (In view of homogienity we may assume this]. Write ax+by+cz as $\alpha (y+z)+\beta (z+x) + \gamma (x+y)$ The coefficients now add up to 2. After mulriplying and dividing by 2 apply convexity of $x \to \frac 1 x$ on $(0,\infty)$ to get an inequality very similar to the one above. The coefficients on the left are now different. $\endgroup$ Dec 14, 2017 at 9:51
  • $\begingroup$ Yes, I tried your method. We can even assume that $x+y+z=1$. And I even tried the limit case $z=0$, which unfortunately holds, too. Hard to find a counterexample, I still tend to believe it is true. $\endgroup$
    – Mathis
    Dec 14, 2017 at 14:18
  • $\begingroup$ The inequality is true. The Buffalo Way works although it is an ugly solution. The inequality is symmetric with respect to $(a, x), (b, y)$ and $(c,z)$. Thus, one may assume that $z \le y \le x$. $\endgroup$
    – River Li
    Oct 5, 2019 at 15:44
  • $\begingroup$ Might be useful: artofproblemsolving.com/community/c6h1688p535460 $\endgroup$ Dec 28, 2019 at 20:45

3 Answers 3


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


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

  • $\begingroup$ Hi Alex can you check my proof with the answer of Gimusi ? $\endgroup$
    – user448747
    Dec 28, 2017 at 19:10
  • $\begingroup$ @FatsWallers Unfortunately, Gimusi’s proof is too simple to be true, see my comment there. $\endgroup$ Dec 29, 2017 at 11:25
  • $\begingroup$ Hi Alex Do you think we can use the Equal Variable Method ? $\endgroup$
    – user448747
    Jan 4, 2018 at 21:25
  • $\begingroup$ @FatsWallers Unfortunately, I didn't hear about this method. $\endgroup$ Jan 5, 2018 at 4:51
  • $\begingroup$ See here emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf .I think it's useful in your case maybe . $\endgroup$
    – user448747
    Jan 5, 2018 at 12:02

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

  • $\begingroup$ Thanks for the code, now I'm more confident about the truth of the inequality. $\frac{1}{b(x+z)}+\frac{1}{c(x+y)}\ge\frac{3}{ax+by+cz}$ doesn't hold: $a=b=c=1$, $z=0$, then it becomes $\frac{1}{x}+\frac{1}{x+y}\geq \frac{3}{x+y}$, or equivalently $y\geq x$. $\endgroup$
    – Mathis
    Dec 18, 2017 at 0:06

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 :




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 :




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 .

  • $\begingroup$ The idea looks good, but I don't know whether the substitution is valid (does such a triangle exist?). Besides, the inequality obtained seems more complicated than the original one. $\endgroup$
    – Mathis
    Dec 17, 2017 at 12:48
  • 3
    $\begingroup$ Let $x=10$, $y=z=1$. Then $p=q=11$, $r=2$. Next, $CA=\sqrt{\frac{pr}q}=\sqrt{2} $, $BC=\sqrt{\frac{qr}p}=\sqrt{2} $, and $AB=\sqrt{\frac{pq}2}=\sqrt{60.5}> 2\sqrt{2}=BC+CA$, so the triangle $ABC$ doesn’t exist. $\endgroup$ Dec 23, 2017 at 11:44
  • $\begingroup$ Hi Alex in fact yes it doesn't work that's why the next idea is to generalize this with the Steiner inellipse and hope it works If you have some good ideas see here . math.stackexchange.com/questions/2571971/steiner-inellipse $\endgroup$
    – user448747
    Dec 23, 2017 at 12:56
  • $\begingroup$ I have a serious way to prove it with this mathweb.scranton.edu/monks/courses/ProblemSolving/… $\endgroup$
    – user448747
    Dec 28, 2017 at 12:54
  • $\begingroup$ @FatsWallers Thanks a lot for the quote! You can also insert the full solution here. $\endgroup$
    – user
    Dec 28, 2017 at 21:40

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