<s>[INST] proving :$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$ Let $a,b,c, >0$ be real numbers such that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3$$
How to prove that :
$$\frac{(ab+b)(2b+1)}{(ab+a)(5b+1)}+\frac{(bc+c)(2c+1)}{(bc+b)(5c+1)}+\frac{(ca+a)(2a+1)}{(ca+c)(5a+1)}\ge\frac{3}{2}$$
 [/INST] I) Let us define for $x\geq 0$ the rational function
$$\tag{1}
f(x)~:=~ \frac{(x+1)(x+5)}{x+2}~=~ x+4 - \frac{3}{x+2}~>~0, \qquad x\geq 0. $$
It is not hard to see that $f$ is positive, monotonically increasing
$$\tag{2}
f'(x)~=~ 1 + \frac{3}{(x+2)^2}~>~0, \qquad x\geq 0, $$
and concave for $x\geq 0$. The tangent at $x=1$ is above the concave function,
$$\tag{3} f(x) ~\leq~ f(1) +(x-1) f'(1)~=~\frac{4}{3}(x+2), \qquad x\geq 0.$$
II) By going to reciprocal variables OP's inequality (v7) can be written as
$$\tag{4}
\forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow\quad \frac{x+1}{f(y)} + \frac{y+1}{f(z)} + \frac{z+1}{f(x)} ~\stackrel{?}{\geq}~ \frac{3}{2}. $$
III) If we could prove  that
$$\tag{5}
\forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow\quad \frac{x+1}{y+2} + \frac{y+1}{z+2} + \frac{z+1}{x+2} ~\stackrel{?}{\geq}~ 2, $$
then eq. (4) would follow from eqs. (3) and (5). 
IV) Equation (5) can be rewritten as
$$\tag{6}
\forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow \quad g(x,y,z)~\stackrel{?}{\geq}~ 0, $$
where
$$ g(x,y,z) ~:=~-2(x+2)(y+2)(z+2)+ \sum_{{\rm cycl.}~ x,y,z}(x+1)(x+2)(y+2) $$ 
$$~=~ -2xyz -4 +\sum_{{\rm cycl.}~ x,y,z}\left(x^2y+2x^2-xy\right)$$ 
$$ \tag{7}  ~=~ 2\underbrace{\left(-xyz+\frac{1}{3}\sum_{{\rm cycl.}~ x,y,z}x^2y\right)}_{\geq 0 ~\text{because of (AG)}.}
+\underbrace{\left(-3+\sum_{{\rm cycl.}~ x,y,z}x^2\right)}_{\geq 0}+h(x,y,z),  $$
where
$$ \tag{8} h(x,y,z) ~:=~ -1+\sum_{{\rm cycl.}~ x,y,z}\left(\frac{1}{3}x^2y+x^2-xy\right). $$
In equation (7) we have used the inequality of arithmetic and geometric means (AG).
V) If we could prove that
$$\tag{9}
\forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 \geq 3 \quad\Rightarrow \quad h(x,y,z)~\stackrel{?}{\geq}~ 0, $$
then eq. (6) would follow from eqs. (7) and (9).
VI) The function $h$ has no internal stationary points because the radial derivative is strictly positive:
$$r \frac{\partial h(x,y,z)}{\partial r}
~=~x \frac{\partial h(x,y,z)}{\partial x}+y \frac{\partial h(x,y,z)}{\partial y}+z \frac{\partial h(x,y,z)}{\partial z}$$
$$\tag{10}~=~\sum_{{\rm cycl.}~ x,y,z}\left(x^2y+2x^2-2xy\right)
 ~=~\sum_{{\rm cycl.}~ x,y,z}\left(x^2y+(x-y)^2\right)~>0.$$
Thus we may restrict to the sphere $x^2+y^2+z^2=3$ from now on. In other words, if we could prove  that
$$\tag{11} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 = 3 \quad\Rightarrow \quad h(x,y,z)~\stackrel{?}{\geq}~ 0, $$
then eq. (9) would follow.
VII) Let us rewrite
$$\tag{12}  h(x,y,z) ~=~ 2 + \underbrace{\left(-3+\sum_{{\rm cycl.}~ x,y,z}x^2\right)}_{= 0}  - j(x,y,z),  $$
where
$$ \tag{13} j(x,y,z) ~:=~ \sum_{{\rm cycl.}~ x,y,z}k(x)y, $$
and where the concave function $k$ is
$$ \tag{14} k(x) ~:=~ x(1-\frac{x}{3}) $$
with derivative $k'(x) = 1-\frac{2x}{3}$. The tangent at $x=1$ is above the concave function,
$$\tag{15} k(x) ~\leq~ k(1) +(x-1) k'(1)~=~\frac{x+1}{3}.$$
If we could prove that
$$\tag{16} \forall x,y,z\geq 0:\quad x^2 +y^2 +z^2 = 3 \quad\Rightarrow \quad j(x,y,z)~\stackrel{?}{\leq}~ 2, $$
then eq. (11) would follow.
VIII)  Proof of eq. (16):
$$\tag{17} j(x,y,z) ~\leq~\frac{1}{3}\sum_{{\rm cycl.}~ x,y,z}(x+1)y ~\leq~2. $$
The first inequality follows from eqs. (13) and (15). The second inequality follows because 
$$\tag{18} \sum_{{\rm cycl.}~ x,y,z}xy ~\leq~ \sum_{{\rm cycl.}~ x,y,z}x^2 ~=~3, $$ 
and because $(x,y,z)=(1,1,1)$ is the maximum point for the function $(x,y,z)\mapsto x+y+z$ on the sphere $x^2+y^2+z^2=3$.
IX) Working backwards through the chain of arguments, we have proven OP's inequality.
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