Stronger than Nesbitt inequality For $x,y,z &gt;0$, 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
Here's a proof.

Using Lagrange's identity leads to 
$$\sum_{cyc}\frac{x}{y+z}=\frac32+\frac12\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \qquad (*)$$

We get this by letting $a = \sqrt{x+y}$, $b = \sqrt{y+z}$, $c = \sqrt{z+x}$, 
$d = \frac{1}{\sqrt{x+y}}$, $e = \frac{1}{\sqrt{y+z}}$, $f = \frac{1}{\sqrt{z+x}}$, and by Lagrange's identity, 
$$(a^2+b^2+c^2)(d^2+e^2+f^2) = (ad+be+cf)^2 + (ae-bd)^2+(af-cd)^2+(bf-ce)^2$$
which gives 
$$2 (x+y+z)(\sum_{cyc}\frac{1}{y+z}) = 9 + \sum_{cyc} \left(\frac{\sqrt{x+y}}{\sqrt{y+z}} -  \frac{\sqrt{y+z}}{\sqrt{x+y}}\right)^2$$
or
$$3 +\sum_{cyc}\frac{x}{y+z} = \frac92 + \frac12\sum_{cyc} \frac{\left({x+y} - (y+z)\right)^2}{({y+z})(x+y)} $$
which is the desired equation $(*)$.

Squaring both sides of the inequality then gives

$$
\tag{1} \frac16 \left[ \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)}  \right ]^2 + \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geq \frac{(y-z)^2}{xy + yz + xz}
$$

We will follow two paths for separate cases.

Path 1:
Omitting the square term   it suffices to prove
$$\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant  \frac{(y-z)^2}{xy + yz + xz}$$

Clearing denominators, we obtain
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq  \frac{(y-z)^2}{xy + yz + xz} (x+y) (y+z) (z+x)
$$

Using $(x+y)(y+z)(x+z) = (x+y+z)(xy + yz+ xz) - x y z$
it suffices to show

$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq  {(y-z)^2} (x+y+z)
$$
or 
$$
(x-y)^2 (x+y) +  (z-x)^2 (z+x) \geq  (y-z)^2 x
$$

Since 
$$
 (y-z)^2 = (y-x + x- z)^2 = (y-x)^2  + (x- z)^2 + 2 (y-x)(x-z) 
$$
this translates into 

$$
(y-x)^2 y +  (x-z)^2 z \geq  2 x(y-x)(x-z) 
$$
For the two cases $y\geq x ; z\geq x $ and $y\leq x  ; z\leq x $ the RHS $\leq 0$ so we are done. For the other two cases, by symmetry, it remains to show the case  $y&gt;  x  ; z &lt;  x $. 

Rearranging terms, we can also write
$$
(y-x)^3 - (x-z)^3 +x ((y -  x) + (z-x))^2 \geq 0
$$

This holds true at least for $(y-x)^3 \geq  (x-z)^3$ or $y+z\geq 2 x$.

So the proof is complete other than for the case $y+z &lt; 2x$ and [ $y&gt;  x  ; z &lt;  x $ or  $z&gt;  x  ; y &lt;  x $ ].

Path 2.

For the remaining case  $y+z &lt; 2 x $ and  [$y&gt;  x ; z &lt;  x$ or  $z&gt;  x  ; y &lt;  x $]  we will follow a different path. Again, by symmetry, we must inspect only  $y+z &lt; 2x$ and  $y&gt;  x &gt; z$.

A remark up front: In the following, some high order polynomials of one variable have to be inspected. MATLAB is used, also for plotting behaviours of these polynomials. There is no case in spending effort for further analytical work on polynomials where their  behaviour is obvious. Still, what follows contains some "ugly" parts.

In the squared version (1) of the inequality, we can use a further inequality which has been proved here:
$$\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}$$

So it suffices to prove

$$
\frac16 \left[ \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2} \right ]^2 + \sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geq \frac{(y-z)^2}{xy + yz + xz}
$$

Some numerical inspection shows immediately that the first term cannot be ommitted.

Clearing some denominators gives

$$
\tag{2} \frac16 \left[ \frac{27}{8}  \right ]^2 \frac{(y-z)^4 (xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4} + \\
 (xy + yz + xz) \sum_{cyc} {(x-y)^2}{(x+y)} - (y-z)^2 (x+y)(y+z) (z+x) \geqslant 0 \quad 
$$

By homogeneity, we set  $y=1+z$. 

The condition $y+z &lt; 2x$ then translates into $1+2z &lt; 2x$, hence we further set $x = z + (1 +q)/2$ where $0\leq q \leq 1$ since also $x = z + (1 +q)/2 &lt; y = 1 +z$. 

Inserting $y=1+z$ and $x = z + (1 +q)/2$ into (2) is straightforward, the result is lengthy (not displayed here).

Let us start with focussing  on the first term in (2), calling that fraction $F$: 
$$
F= \frac{(y-z)^4 (xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4} 
$$
With the setting $y=1+z$ this  can be simplified to
$$
F  = \frac{(xy + yz + xz) (x+y)(y+z) (z+x)}{(x+y+z)^4} 
$$

Since by the settings $y=1+z$ and $x = z + (1 +q)/2$ both x and y are linear in z,  the numerator of $F$ is of fifth order in $z$, whereas the denominator is of fourth order in $z$. In leading order, the whole term is therefore of first order in $z$ and will therefore rise with $z$ for large enough $z$. This motivates to show that  indeed
 $$F(q,z) \geq F(q,z=0) = 2 \frac{(q+1)^2}{(q+3)^2}$$
for all $z$ and $q$. 
Showing  that directly requires a condition 

$$
G = (xy + yz + xz) (x+y)(y+z) (z+x) (q+3)^2 - 2 (q+1)^2 (x+y+z)^4  \geq 0
$$

Inserting $y=1+z$ and $x = z + (1 +q)/2$ into $G$, and expanding  the brackets,  gives a very lengthy expression which however contains only positive terms, so the condition is proved immediately:

$$
G = (z(2q^6z + 2q^6 + 22q^5z^2 + 51q^5z + 22q^5 + 80q^4z^3 + 358q^4z^2 + 357q^4z + 100q^4 + 96q^3z^4 + 960q^3z^3 + 1828q^3z^2 + 1206q^3z + 260q^3 + 864q^2z^4 + 3672q^2z^3 + 4788q^2z^2 + 2484q^2z + 450q^2 + 2592qz^4 + 7344qz^3 + 7398qz^2 + 3159qz + 486q + 2592z^4 + 5832z^3 + 4806z^2 + 1701z + 216))/4 \geq 0
$$

Hence it suffices, instead of (2), to prove the following:

$$
\tag{3} \frac16 \left[ \frac{27}{8}  \right ]^2 2 \frac{(q+1)^2}{(q+3)^2} + 
 (xy + yz + xz) \sum_{cyc} {(x-y)^2}{(x+y)} \\ -  (x+y)(y+z) (z+x) \geqslant 0 \quad 
$$

After insertion of $y=1+z$ and $x = z + (1 +q)/2$, the factors $(x-y)^2$ in the cyclic sum will not be functions of $z$.  Hence,
the LHS  is a third order expression in $z$ with leading (in $z^3$ ) term $( 1 + 3 q^2) z^3$, so for large enough $z$ it is rising with $z$. A remarkable   feature of this expression is that for the considered range $0\leq q \leq 1$ it is actually monotonously rising for all $z$. To see this, consider whether there are points with zero slope. The first derivative of the expression with respect to $z$ is 

$$
q^4/4 + (7q^3z)/2 + (7q^3)/4 + 9q^2z^2 + 9q^2z + (5q^2)/4 - (7qz)/2 - (7q)/4 + 3z^2 + 3z + 1/2
$$
Equating this to zero gives
$$
z_{1} = 
-(2((13q^6)/4 + (17q^4)/2 + (133q^2)/4 + 3)^{(1/2)} - 7q + 18q^2 + 7q^3 + 6)/(36q^2 + 12)\\
z_{2} = 
-(-2((13q^6)/4 + (17q^4)/2 + (133q^2)/4 + 3)^{(1/2)} - 7q + 18q^2 + 7q^3 + 6)/(36q^2 + 12)
$$
One now shows that in the range $0\leq q \leq 1$, there are only negative solutions $z_{1,2}$. So we will have no zero slopes for $z\geq 0$. Since the polynomials  are (roots of)  sixth order in q, we investigate the following figures:







So monotonicity (rising with $z$) is established for all $q$.

Hence, to show the inequality it suffices to inspect  (3) at the smallest $z=0$. This gives 

$$
((q + 1)(8q^6 + 88q^5 + 336q^4 + 432q^3 - 216q^2 - 405q + 243))/(64(q + 3)^3) \geqslant 0 \quad 
$$
or

$$
\tag{4} 8q^6 + 88q^5 + 336q^4 + 432q^3 - 216q^2 - 405q + 243 \geqslant 0 \quad 
$$

An even stronger requirement is 
$$
h(q) = 
432q^3 - 216q^2 - 405q + 243 \geq 0
$$
In the considered range $0\leq q \leq 1$, $h(q)$ has a minimum which is obtained by taking the first derivative,
$$
1296 q^2 - 432 q - 405
$$
and equating to zero, which gives $q = 3/4$, and the above $h(q)$ then gives
$$
h(q = 3/4) = 0
$$

This establishes the inequality. $ \qquad \Box$
Very long partial answer :
As in my other answer we have using abc method or uvw method the inequality :
Let $a,b,c&gt;0$
$$\sum_{cyc}\frac{a}{b+c}\geq \sqrt{\frac{9}{4}+\frac{7}{4}\frac{\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)}{ab+bc+ca}}$$
See WA factorization for the case wich is sufficient to show I mean ,$b=c=1$ (because it's homogenous).
As partial answer  we have the constraint :
$$\frac{7*2}{4*3}(a^{2}+b^{2}+c^{2}-ab-bc-ca)\geq (a-b)^2$$
Hope it inspires you !
Update 22/04/2021
Let $a,b,c\in[0.5,1]$ then we have :
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+0.5\frac{(a-b)^2}{ab+bc+ca}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
Proof :
The RHS is equivalent to :
$$0.5\frac{(a-b)^2}{ab+bc+ca}\geq 0$$
For the LHS we can use BW easily .
We can go further as we have :
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+0.5\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)} \geq 1.5+0.5\frac{(a-b)^2}{ab+bc+ca}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
With the constraint :
$$(0.5(a+b-2c)(a^2-ab+b^2-c^2))\geq 0$$
Again I go further :
We have
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+\frac{44}{100}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)} \geq \sqrt{\frac{9}{4}+\frac{132}{100}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}} \geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
With the first constraint :
$$ (28 a^3 - 3 a^2 b - 47 a^2 c - 3 a b^2 + 44 a b c - 25 a c^2 + 28 b^3 - 47 b^2 c - 25 b c^2 + 50 c^3)\geq 0$$
And the second :
$$\frac{132}{100}\frac{2}{3}(a+b+c)(ab+bc+ca)\geq(a+b)(b+c)(c+a)$$
Update 23/04/2021 :
A general way would be :
We have
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+y\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)} \geq \sqrt{\frac{9}{4}+3y\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}} \geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}\quad(I)$$
With the constraint :
$$-(0.5 (2 a^3 y - 2 a^3 - 2 a^2 b y + a^2 b + 2 a^2 c y + a^2 c - 2 a b^2 y + a b^2 - 4 a b c y + a c^2 + 2 b^3 y - 2 b^3 + 2 b^2 c y + b^2 c + b c^2 - 2 c^3))\geq 0$$
And :$$3y\frac{2}{3}(a+b+c)(ab+bc+ca)\geq(a+b)(b+c)(c+a)$$
So to be true we have :
$$\frac{\left(a\left(a+b\right)\left(a+c\right)+b\left(b+c\right)\left(b+a\right)+c\left(a+c\right)\left(c+b\right)-1.5\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)}{\left(a-b\right)^{2}\left(a+b+c\right)}-\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(a+b+c\right)\left(ab+bc+ca\right)}\geq 0$$
We have to establish an assumption because like this the inequality is false ($a,b,c&gt;0$ is not sufficient)
A temporary and sufficient assumption is $a\in[0.25,0.5)$ and $b,c\in[0.5,1]$
Using the first inequality and using the variable $y$ as in the inequality $I$
We need to show for $a\ge c\ge b$ and $2c-2b\le a$ :
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)-\frac{\left(a-b\right)^{2}}{2}\geq 0\quad (C)$$
Update 24/04/2021
We have the inequality for  $1\ge a\geq c\geq b&gt;0$ and $\frac{\left(a-b\right)^{2}}{ab+bc+ca}\le 1$ or $\frac{\left(a-b\right)^{2}}{ab+bc+ca}\ge 1.2$ and $2c-2b\ge a$:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge 1.5+\frac{\left(\sqrt{\frac{15}{4}}-1.5\right)\left(\left|a-b\right|\right)^{1.75}}{\left(ab+bc+ca\right)^{\frac{1.75}{2}}}\geq  \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}} $$
Update 25/04/2021
To prove it we remark :
$$f(c)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5$$
Is increasing with $a\leq 2c-2b$
$$g(c)=\left(ab+bc+ca\right)^{\frac{1.75}{2}}$$
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 .
Update 26/04/2021 :
To prove partialy the inequality $(C)$ we have for $a\ge 2c\ge c\ge b$ and $2c-2b\le a$  :
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)-\frac{\left(a-b\right)^{2}}{2}\geq \frac{2(a^2+b^2+c^2-ab-bc-ca)}{3}\geq\frac{\left(a-b\right)^{2}}{2}\quad $$
Wich is trivial using the $uvw's$ method and the assumptions on $a,b,c&gt;0$
For the case $2c\geq a \geq c\geq b$ we add the assumptions $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 too hard i think .
From my first answer we have :
$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Remains to show :
$$P(a,b,c)\geq \sqrt{\frac{9}{4}+\frac{3}{2}\frac{\left(a-b\right)^2}{ab+bc+ca}}$$
With assumptions  $2c\geq a \geq c\geq b$ and $c+b\geq a $ and $a\geq 2c-2b$ .
We have :
Proof : we have the refinement of the inequality above :
$$\frac{9}{1}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}\geq \frac{9}{4}\frac{\left(a-b\right)^4}{(ab+bc+ca)^2}$$
Multiplying by the differents denominators and replacing $a$ by $c+b$ and $a$ by $2c-2b$ . Eliminating a variable because the inequality is homogenous we get a decreasing polynomial for the LHS and an increasing polynomial for the RHS wich are :
$$4((2-2x)(1+x)+x)^2(2+x)((2-2x)^4+1+x^4-(x^2+1)(2-2x)^2)((2-2x)^2+1+x^2-(x+1)(2-2x))\geq ((x+1)(x+2)(1+2x))^2(0.75(2+2x)+0.75((1+x)x)^{\frac{1}{3}})$$
With $0.33\leq x\leq 0.3675$ and $x=\frac{b}{c}$
Hope it helps you !