Stronger than Nesbitt's inequality $\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$ 
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
  $$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$$

Nesbitt's inequality is the following:
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2},$$
which follows from C-S:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$
but this way does not help for the starting inequality.
There is a nice solution for the following inequality, which was in our test six months ago. 
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
By C-S $$\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{(b+c)^2}{ab+ac+2bc}.$$
Thus, it's enough to prove that
$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{(b+c)^2}{ab+ac+2bc}\geq\frac{3}{2}$$ or
$$2(b+c)a^2-(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc)a+2(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq0.$$
Thus, it's enough to prove that
$$16(b+c)(b^2-bc+c^2)\sqrt[3]{4(b^3+c^3)}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ or
$$16\sqrt[3]{4(b^3+c^3)^4}\geq\left(3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc\right)^2$$ and since
$$3(b+c)\sqrt[3]{4(b^3+c^3)}>4bc,$$
it remains to prove that
$$4\sqrt[3]{2(b^3+c^3)^2}\geq3(b+c)\sqrt[3]{4(b^3+c^3)}-4bc$$ and after using
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz),$$ where
$$x^2+y^2+z^2-xy-xz-yz\neq0$$ we need to prove that
$$128(b^3+c^3)^2-108(b+c)^3(b^3+c^3)+64b^3c^3+288(b+c)(b^3+c^3)bc\geq0.$$
Now, let $b^2+c^2=2kbc$.
Hence, we need to prove that:
$$128(2k+2)(2k-1)^2-108(2k+2)^2(2k-1)+64+288(2k+2)(2k-1)\geq0$$ or
$$(k-1)^2(10k+11)\geq0.$$
Done!
But this way gives a wrong inequality again. 
 A: Your inequality is equivalent to :
$$f(x)=\frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}$$
Here we assume that $x\geq h \geq 1$
It's easy to see that the function $f(x)$ is convex with $h$ fixed (as the sum of convexs functions) and increasing for $x\geq h$
Furthermore for $h$ fixed we have that the minimum is inferior to $h$ (in abscissa)
So we have :
$$f(x)\geq f(h)$$
We get :
$$f(x)=\frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq\frac{1}{(8(h^4+h^4))^{0.25}}+\frac{h}{h+1}+\frac{h}{h+1} $$
So the inequality is verified in this case because it's easy to threat this one variable inequality. 
The case where $h\geq x \geq 1$ is the same (because of the symetry) and the cases $x\geq 1 \geq h$ and  
$h\geq 1 \geq x$ are similar for the same reasons 
Now we continue with the case $x\leq h \leq 1$
Remark that if we inverse each variable it becomes :
$$\frac{xh}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}$$
With $h\geq 1$ and $x\geq 1$ but we have :
$$\frac{xh}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq \frac{1}{(8(x^4+h^4))^{0.25}}+\frac{x}{h+1}+\frac{h}{x+1}\geq 1.5$$
So this case is verified .
Done !
Edit :My proof is incomplete like this because the case where  $x\geq 1 \geq h$ is not solved . So I purpose to you a solution :
The main idea is to use the three chord lemma related to the convexity of the function $f(x)$ so we have for $x\geq \sqrt{x}\geq x^{0.25}$ :
$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq \frac{f(x)-f(x^{0.25})}{x-x^{0.25}} \geq \frac{f(\sqrt{x})-f(x^{0.25})}{\sqrt{x}-x^{0.25}}$$
We can apply this inequality $n$ times to get :
$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq \frac{f(x^{1/2^n})-f(x^{1/2^{n+1}})}{x^{1/2^n}-x^{1/2^{n+1}}}$$ 
But we have to prove that the RHS is positive for a variable wich tends to $1$ so we get the following result to evaluate :
$$\lim_{x\to 1} \frac{\dfrac{1}{(8(x^{1/2^n}+h^4))^{0.25}}+\dfrac{x^{1/2^n}}{h+1} + \dfrac{h}{x^{1/2^n}+1}- \left(\dfrac{1}{(8(x^{1/2^{n+1}}+h^4))^{0.25}} + \dfrac{x^{1/2^{n+1}}}{h+1}+\dfrac{h}{x^{1/2^{n+1}}+1} \right)}{x^{1/2^n}-x^{1/2^{n+1}}}$$
For that see the the interesting answer of G.Cab here
With this answer we know that the result does not depends on $n$ and is equal to :
$$
\eqalign{
  & \mathop { }
  &   {1 \over {h + 1}} - {1 \over {2^{\,11/4} \left( {h^{\,4}  + 1} \right)^{\,5/4} }} - {h \over 4} \cr} 
$$
Quantity wich is positive for $h\leq 1$
So we get :
$$\frac{f(x)-f(\sqrt{x})}{x-\sqrt{x}}\geq 0$$
Or :
$$f(x)\geq f(\sqrt{x})$$
If we apply this inequality $n$ times we get :
$$f(x)\geq f(x^{\frac{1}{2^n}})$$
But the variable tends $1$ and the RHS to $1.5$ so we get the desired result !
