Stronger than Nesbitt inequality 
For $x,y,z >0$, prove that
  $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \sqrt{\frac94+\frac32 \cdot \frac{(y-z)^2}{xy+yz+zx}}$$

Observation:


*

*This inequality is stronger than the famous Nesbitt's Inequality
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32 $$ for positive $x,y,z$

*We have three variables but the symmetry holds only for two variables $y,z$, resulting in a very difficult inequality. Brute force and Largrange Multiplier are too complicated.

*The constant $\frac32$ is closed to the best constant. Thus, this inequality is very sharp, simple AM-GM estimation did not work.


Update: As point out by Michael Rozenberg, this inequality is still unsolved

Abstract :
We suggests a way to show the OP inequality so some part are left to the reader because there are easy  . We spilt the problem in two  cases , first case $2c-2b\le a$ and the second $2c-2b\ge a$ . As the problem is homogenous we suppose in addition $1\ge a\ge c\ge b$.Following that we use  convexity and derivatives.

First case : $1\ge a\ge c\ge b$ and $2c-2b\le a$
we have :
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+0.5\frac{(a-b)^2}{ab+bc+ca}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
Or:
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)-\frac{\left(a-b\right)^{2}}{2}\geq 0\quad (C)$$

SubCase : $a\ge 2c\ge c\ge b$
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)\geq \frac{2(a^2+b^2+c^2-ab-bc-ca)}{3}\geq\frac{\left(a-b\right)^{2}}{2}\quad $$
Wich is trivial using the $uvw's$ method and the assumptions on $a,b,c>0$

Subcase: $2c\geq a \geq c\geq b$ and $c+b\leq a $ and $a\geq 2c-2b$ remarking that :
$$f(c)=\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)$$
Is a decreasing function always with the assumptions above .Remains to show the cases $a=c+b$ or $a=2c-2b$ wich is a one variable inequality since all the equalities/inequalities are homogenous.I shall prove that the function is decreasing later .
To prove that the function is decreasing we differentiate twice we have :
$$f''(c)=-\frac{2a^2b}{(a+c)^3}-\frac{2ab^2}{(b+c)^3 }+2$$
With the assumptions above the function $f(c)$ is convex so the derivative is increasing .Remains to show the inequality at the equality case wich is not  hard .

Second case: $1\ge a\geq c\geq b>0$ and  $2c-2b\ge a$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge 1.5+\frac{\left(\sqrt{\frac{15}{4}}-1.5\right)\left(\left|a-b\right|\right)^{1.75}}{\left(ab+bc+ca\right)^{\frac{1.75}{2}}}\geq  \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}} $$
To prove it we remark :
$$f(c)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5$$
Is increasing with $a\leq 2c-2b$ as $f(c)$ is a convex function we deduce that the first derivative is increasing . Remains to replace by the constraint $2c-2b=a$
$$g(c)=\left(ab+bc+ca\right)^{\frac{1.75}{2}}$$
Is increasing .
So the product of two positives increasing functions is also an increasing function always with constraints above .
It remains to show the case $a=2c-2b$ . As it's homogenous we get an inequality with one variable or a long polynomial .The RHS is trivial .
Last Edit 09/03/2022 :
We have the inequality :
Let $0\leq x\leq 1$ then we have :
$$\sqrt{\frac{9}{4}+1.5x}\leq g(x)=\left(2-a\right)\frac{1}{2}\sqrt{\frac{3}{5}}\left(ax-a\right)+\sqrt{\frac{9}{4}+1.5}+\frac{-\sqrt{\frac{3}{5}}}{20}\left(a^{-1}x-a^{-1}\right)^{2}$$
Where $a\geq 0$ is chosen as :
$$g(0)=1.5$$
We can do really better as it seems we have for $0\leq x\leq 1$ :
$$\left(2-a^{1.0735}\right)\frac{1}{2}\sqrt{\frac{3}{5}}\left(ax-a\right)+\sqrt{\frac{9}{4}+1.5}+\frac{-\sqrt{\frac{3}{5}}}{20}\left(a^{-1}x-a^{-1}\right)^{2}\geq \sqrt{\frac{9}{4}+1.5x}$$
Where $a\simeq 0.7748951014$ is chosen as :
$$g(0)=1.5$$
