Lemma: [Smoothing of convex functions] Given a convex function $f(x)$, and variables $ a \leq b$, if $ \epsilon < b-a$, then $ f(a) + f(b) \geq f( a + \epsilon ) + f(b- \epsilon)$.
(This is "well-known", so I'm not going to prove it. If you're stuck, explain what you've tried.)
Corollary: Given a convex function $f(x)$, and variables $ a \leq b \leq c$, $ A \leq B \leq C$ such that $ a + b + c = A + B + C $, $ C \geq c$, and $A \leq a$, then $ f(A) + f(B) + f(C) \geq f(a) + f(b) + f(c)$.
Comments:
The proof follows by applying the above lemma (or see the hidden block).
I consider this "less well-known", though it's still often used for Olympiad inequalities.
The 4-variable analogue need not hold, because we don't have enough control over the middle terms.
I (personally) call this the "3-variable smoothing Jensens".
Proof: Let $ C - c = c' $, $ A - a = - a' $, $B - b = a' - c' $.
If $B - b > 0$, then $ f(B) + f(A) \geq f(b) + f(A+B-b)$ by the lemma.
We now have $ C + (A+B-b) = c + a$, so $ f(C) + f(A+B-b) \geq f(c) + f(a) $ by the lemma.
Thus, $ f(A) + f(B) + f(C) \geq f(a) + f(b) + f(c)$.
Likewise if $ B - b < 0$, then work with $ f(B) + f(C) \geq f(b) + f(B+C - b)$ first.
Corollary: The problem follows.
First, a simplification: To show that the function is increasing, we just need to show that for $ z = x/y > 1$, $$\sum \frac{ \frac{ p^z } {p^z + q^z + r^z } } { 1 - \frac{ p^z } {p^z + q^z + r^z } } \geq \frac{\frac{ p } { p + q + r }}{1 - \frac{ p } { p + q + r } } \quad (1) $$
then apply the substitution $ p = a^x, q = b^x, r = c^x$ to conclude that $f(y) \geq f(x)$.
Now, WLOG $ p = \max (p, q, r)$.
$ \frac{ p^z } {p^z + q^z + r^z } \geq \frac{ p } { p + q + r } \Leftrightarrow p^z q + p^z r \geq p q^z + p r^z$, which is true.
Likewise, WLOG $ r = \min (p,q,r)$ and $ \frac{ r^z } {p^z + q^z + r^z } \leq \frac{ r } { p + q + r }$
Hence, we can apply the previous corollary with $ f(x) = \frac{x}{1-x}$, $\{A, B, C \} = \{ \frac{ p^z } {p^z + q^z + r^z }, \frac{ q^z } {p^z + q^z + r^z }, \frac{ r^z } {p^z + q^z + r^z } \} $, $ \{ a, b, c \} = \{ \frac{ p } { p + q + r }, \frac{ q } { p + q + r }, \frac{ r } { p + q + r }\}$, so inequality (1) is true.
Best Answer
There are very many refinements of the Nesbitt's inequality.
For example. For positives $a$, $b$ and $c$ we have:
1.$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2};$$ 2.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)};$$ 3.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3\sqrt{5(a^2+b^2+c^2)-ab-ac-bc}}{4(a+b+c)};$$ 4. For any reals $a$, $b$ and $c$ such that $ab+ac+bc>0$ prove that: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$$