Elegant but pugnacious inequality 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$
$\frac{1}{u}=c$$\quad$$\frac{1}{p}=x$
$\frac{1}{q}=y$$\quad$$\frac{1}{r}=z$
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.
 A: 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. 
A: 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 . 
