Stronger than Nesbitt's inequality using convexity and functions Hi it's a refinement of Nesbitt's inequality  and for that, we introduce the function :
$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$
With $a,b,x>0$
Due to homogeneity we assume $a+b=1$ and we introduce the function :
$$g(a)=\frac{a}{1-a+x}$$
Showing that $g(a)$ is convex on $(0,1)$ is not hard so we have :
$$\frac{b}{x+a}+\frac{a}{b+x}\geq 2\frac{a+b}{2(\frac{a+b}{2}+x)}=h(x)$$
So we have :
$$f(x)\geq h(x)+\frac{x}{a+b}$$
Now we put $u=\frac{x}{a+b}$ and we want to show :
$$h(x)+\frac{x}{a+b}=u+\frac{1}{0.5+u}\geq \frac{3}{2}$$
The last inequality is obvious.
My question :
It is correct?
Do you know other refinements?
Thanks in advance!
Ps: I add the tag reference request for the last question.
 A: 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:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$$
A: Stronger than Nesbitts inequality for non-cyclic style

*

*If $a,\,b,\,c$ are positive real numbers, then
$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$

*If $a,\,b,\,c$ are positive real numbers, then
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{27(b-c)^{2}}{16(a+b+c)^{2}}.$$

*Let $a,\,b,\,c$ be nonnegative real numbers, no two of which are zero. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{(b-c)^{2}}{2(b+c)^{2}}.$$

*Let $a,b,c$ are non-negative real numbers such that $ab+bc+ca>0.$ Prove that
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \geqslant \dfrac{3}{2}+\dfrac{7(b-c)^2}{16a(b+c)+7bc}.$$
