How prove this inequality: $\frac1{1-a}+\frac1{1-b}+\frac1{1-c}\ge \frac1{ab+bc+ac}+\frac1{2(a^2+b^2+c^2)}$ for $a+b+c=1$? Let $a,b,c>0$ and such $a+b+c=1$ show that 
$$\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}\ge \dfrac{1}{ab+bc+ac}+\dfrac{1}{2(a^2+b^2+c^2)}$$
Let $p=a+b+c=1,ab+bc+ac=q,abc=r$
$$\Longleftrightarrow -4q^3+q^2-3qr+2r\ge 0$$
it seem hard to prove.
why I say it hard prove:
use Schur inequality
$$p^3-4pq+9r\ge 0\Longrightarrow r\ge\dfrac{4q-1}{9}$$
it  remains to prove that
$$\dfrac{4q-1}{9}(2-3q)+q^2-4q^3\ge 0$$
In fact, this inequality can't hold (4q-1)

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $\frac{9u^2+3v^2}{9uv^2-w^3}\geq3u\left(\frac{1}{3v^2}+\frac{1}{18u^2-12v^2}\right)$, 
which is equivalent to $f(w^3)\geq0$, where $f$ is a linear function.
But the linear function gets a minimal value for an extremal value of $w^3$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or
$w^3=x^3-3ux^2+3v^2x$ and we see that the line $y=w^3$ and graph of $y=x^3-3ux^2+3v^2x$ have three common points (draw it!). 
Thus, an extremal value of $w^3$ we get for equality case of two variables
and we need to check also the case $w^3\rightarrow0^+$. 


*

*$b=c$. After homogenization we can assume $b=c=1$, which gives $a(a-1)^2\geq0$;

*$w^3\rightarrow0^+$. 
Let $c\rightarrow0^+$. After homogenization we can assume $b=1$, 
which gives $(a-1)^2\geq0$. Done!
