I have a solid and difficult problem that I can't solve, this is the following :

Let $p,q,r,s,t,u$ be real positive numbers then we have : $$\frac{1}{s}+\frac{1}{t}+\frac{1}{u}+\frac{-3}{\frac{p+r+q}{2}-s-t-u}\geq \frac{1}{p}+\frac{1}{q}+\frac{1}{r} $$

With the condition : $$\frac{s}{p}+\frac{t}{q}+\frac{u}{r}=1 $$

My geometric try :

We know this :

Let ABC be a triangle, and let P, Q, R be any points in the plane distinct from A; B; C; respectively and suppose the cevians AP; BQ; CR meet at T then we have :

$$\frac{TQ}{AQ}+\frac{TP}{BP}+\frac{TR}{CR}=1 $$

So it's a geometric interpretation of our condition .

Now put the following substitution : $\frac{1}{s}=a$$\quad$$\frac{1}{t}=b$



We get :

$$a+b+c+\frac{-3}{\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}-\frac{1}{x}-\frac{1}{y}-\frac{1}{z}}\geq x+y+z $$

Furthermore we have for an interior point (in $ABC$) $P$, the Barrow's inequality .Finally we remark that the inequality above seems to have the behavior of the Barrow's inequality .

Edit : As point out in the comment of Doyun Nam I add the implicit condition $p>q+r$$\quad$$q>p+r$$\quad$$r>p+r$.Thanks to him.

After that I have no more idea...Thanks a lot.

  • $\begingroup$ I've revised many times the solution by vectors but I think that now it works, let me know what you think about it! $\endgroup$
    – user
    Dec 28, 2017 at 17:05
  • $\begingroup$ I think you should add the condition that $$p+q+r\neq2(s+t+u)$$. @DoyunNam has given an example of this. $\endgroup$
    – TheSimpliFire
    Jan 5, 2018 at 10:34

2 Answers 2


If $p=30$, $q=r=5$, and $s=18$, $t=u=1$, then the condition is satisfied, but $p+q+r-2(s+t+u)=0$. Furthermore, for sufficiently small and positive $\epsilon$,

let $p=30$, $q=5$, $r=5$,

and $s=18-6\epsilon$, $t=1+\epsilon$, $u=1$.

Then $\frac{s}{p}+\frac{t}{q}+\frac{u}{r}=1$.

However $$\frac{p+q+r}{2}-s-t-u=5\epsilon,$$ thus

$$\frac{-3}{\frac{p+q+r}{2}-s-t-u} = -\frac{3}{5\epsilon}.$$

If positive $\epsilon$ approaches to $0$, then the above term goes to $-\infty$.

It makes a contradiction.

  • $\begingroup$ First thanks for your interest .It's strange because it implies that this inequality math.stackexchange.com/questions/2564669/… is false .I continue to search for ambiguity . Have a good day . $\endgroup$
    – user448747
    Jan 5, 2018 at 14:05
  • $\begingroup$ I read your answer of math.stackexchange.com/questions/2564669/…. In your answer, I understand that $u$, $v_1$, $v_2$ are vectors, and $uv_2$ and $uv_1$ are dot product. However, $(u v_2)(u v_1) \neq u^2 v_2 v_1$. $\endgroup$
    – Doyun Nam
    Jan 5, 2018 at 15:02
  • $\begingroup$ And in your answer (math.stackexchange.com/questions/2564669/…) you put $p=x+y, q=z+x, r=y+z$ for positive $x,y,z$, I think the condition $p+q>r$,$q+r>p$, and $r+p>q$ should be added. $\endgroup$
    – Doyun Nam
    Jan 5, 2018 at 15:23
  • $\begingroup$ Thanks you (+1) , I add it just now . If you have other idea you are welcome ! $\endgroup$
    – user448747
    Jan 5, 2018 at 17:27

Too long for a comment so :

Your inequality is equivalent to : $$\frac{p-s}{ps}+\frac{t-q}{tq}+\frac{r-u}{ru}\geq \frac{3}{0.5(p+q+r)-s-t-u}$$ So we can apply the corollary 2.3 from this link we get : $$\frac{p-s}{ps}+\frac{t-q}{tq}+\frac{r-u}{ru}\geq \frac{(p+q+r-s-t-u)^{-m+1}}{((p-s)(ps)^{\frac{1}{-m}}+(t-q)(tq)^{\frac{1}{-m}}+(r-u)(ru)^{\frac{1}{-m}})^{-m}}$$

So we have to prove :

$$\frac{3}{0.5(p+q+r)-s-t-u}\leq \frac{(p+q+r-s-t-u)^{-m+1}}{((p-s)(ps)^{\frac{1}{-m}}+(t-q)(tq)^{\frac{1}{-m}}+(r-u)(ru)^{\frac{1}{-m}})^{-m}}$$

We put : $$m=\frac{ln(0.5(p+q+r)-s-t-u)}{ln(p+q+r-s-t-u)}+1$$

So we have : $$ (p+q+r-s-t-u)^{-m+1}=\frac{1}{0.5(p+q+r)-s-t-u}$$

So we get :

$$3\leq \frac{1}{((p-s)(ps)^{\frac{1}{-m}}+(t-q)(tq)^{\frac{1}{-m}}+(r-u)(ru)^{\frac{1}{-m}})^{-m}}$$

Or: $$3\leq ((p-s)\frac{1}{(ps)^{\frac{1}{m}}}+(t-q)\frac{1}{(tq)^{\frac{1}{m}}}+(r-u)\frac{1}{(ru)^{\frac{1}{m}}})^m$$

After that the idea is to use Holder's inequality but I'm stuck .Maybe it could help .


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