A difficult symmetric inequality In my studies of various geometric inequalities I reached an inequality which seems true (numerically) but I cannot prove it. Let $p$, $q$, and $r$ be real numbers from the interval $(0,1)$. Let's also define the following function $$f({p})=\frac{\sqrt{1-p}}{(2-p)^2}$$ Prove (or disprove) that: $$
\frac{f(p)+f(q)+f(r)}{\sqrt{p q r}}\leq \frac{f(p)}{p\sqrt{p}}+\frac{f(q)}{q\sqrt{q}}+\frac{f(r)}{r\sqrt{r}}
$$
I've tried Lagrange multipliers but the resulting equations do not seem tractable.
EDIT:
The original question had the condition $p+q+r=2$ which apparently is not necessary, so I dropped it. I can prove that the inequality holds for $p=q$. A possible strategy is to try to establish monotonicity in one of the parameters under certain conditions. Unfortunately I can't manage the calculations.
 A: I was able to prove this, finally. Here is a brief sketch of the proof. I will use the following simple fact:
Lemma. For positive numbers, if $a\geq b\geq c$ and $(x_1,x_2,x_3)\succ(y_1,y_2,y_3)$ then $ax_1+bx_2+cx_3\geq ay_1+by_2+cy_3\geq ay_i+by_j+cy_k$ where $(i,j,k)$ is an arbitrary permutation of $(1,2,3)$
Now notice that the function : $g(p)=f(p)/\sqrt{p}$ is decreasing in $(0,1)$. Assume $p\leq q\leq r$. Our inequality is equivalent to:$$\frac{g(p)}{p}+\frac{g(q)}{q}+\frac{g(r)}{r}\geq\frac{g(p)}{\sqrt{q r}}+\frac{g(q)}{\sqrt{p r}}+\frac{g(r)}{\sqrt{p q}}$$ Let's put $x_1=1/p, x_2=1/q$, $x_3=1/r$ and $y_1=(x_1+x_2)/2, y_2=(x_1+x_3)/2, y_3=(x_2+x_3)/2$. Notice that $x_1\geq x_2\geq x_3$, $y_1\geq y_2\geq y_3$ and $(x_1,x_2,x_3)\succ(y_1,y_2,y_3)$. Applying the lemma for $a=g(p), b=g(q)$ and $c=g(r)$ ($a\geq b\geq c$ because $g(x)$ is decreasing) we get:
$$
ax_1+bx_2+cx_3\geq ay_3+by_2+cy_1=a\frac{x_2+x_3}{2}+b\frac{x_1+x_3}{2}+c\frac{x_1+x_2}{2}\geq a\sqrt{x_2 x_3}+b\sqrt{x_1 x_3} + c\sqrt{x_1 x_2}
$$
and this is exactly what we are trying to prove.
