Given is equivalent το $\displaystyle \sum \dfrac{(a+b+c)-a^2}{a(a+b+c)+bc}\geq 6$. $\displaystyle \sum \dfrac{(a+b+c)-a^2}{a(a+b+c)+bc}=\sum \dfrac{(b+c)(a+c+a+b)}{(a+b)(a+c)}$ Set $a+b=x,b+c=y,a+c=z$.We have to show that $\displaystyle \sum \dfrac{y(x+z)}{xz}\geq 6$. This is true because $\displaystyle \sum \dfrac{y(x+z)}{xz}=\sum (\dfrac{y}{z}+\dfrac{y}{x})=\sum (\dfrac{y}{z}+\dfrac{z}{y})\geq 3\cdot 2=6$.QED

Using the given condition, we have $\frac{1-a^2}{a+bc}=\frac{(1-a)(1+a)}{a(a+b+c)+bc}=\frac{(b+c)(2a+b+c)}{(a+b)(c+a)}=\frac{b+c}{a+b}+\frac{b+c}{c+a}$. Therefore our inequality becomes $$\frac{a+b}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+\frac{c+a}{b+c}\geq 6$$which follows from straight-forward AM-GM for the $6$ terms in the $LHS$. Equality is reached if and only if $a=b=c=\frac{1}{3}$.

$$\frac{1-a^2}{a+bc}==\frac{b+c}{a+b}+\frac{b+c}{c+a}$$

Let $a, b, c$ be positive real numbers such that $a + b^2 + c^2= 1$. Show that $$\frac{1 - a^2}{a + bc} + \frac{1 - b^2}{b + ca} + \frac{1 - c^2}{c + ab} \ge 3$$Let $a, b, c\geq 0,a + bc= 1$ and $ab + bc+ca\neq0 .$ Show that $$\frac{1 - a^2}{a + bc} + \frac{1 - b^2}{b + ca} + \frac{1 - c^2}{c + ab} \ge 1$$

Just write $a+b+c$ when you see $1$.We write $b+c$ and $a+b+a+c$ instead of $1-a$and $1+a$, respectively. Then write $1-b-c$ [for factorizing $bc-b-c+1$] instead of $a$. So we get that $\sum \dfrac{(b+c)(a+c+a+b)}{(a+b)(a+c)}$=$\frac{b+c}{a+b}+\frac{b+c}{c+a}$ by easy $AM-GM$ get desired result. The equality holds if and only if $a=b=c=\frac{1}{3}$

Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Show that $$ \frac{1 - 5a^2}{a + bc} + \frac{1 - 5b^2}{b + ca} + \frac{1 - 5c^2}{c + ab} \ge 3$$$$\frac{1 - a^2}{a + 2bc} + \frac{1 - b^2}{b + 2ca} + \frac{1 - c^2}{c + 2ab} \ge \frac{24}{5}$$

We have \[ \frac{1-a^2}{a+bc} = \frac{(1-a)(1+a)}{a+bc} = \frac{(b+c)(1+a)}{a+bc} = \frac{b+ab+c+ac}{a+bc} \]Thus, it suffices to prove \[ \sum \left(\frac{b+ab+c+ac}{a+bc}+1\right) = (a+b+c+ab+bc+ca)\sum \frac{1}{a+bc}\ge 9, \]which is immediate by AM-HM, as \[ \sum \frac{1}{a+bc}\ge \frac{9}{a+b+c+ab+bc+ca}. \]

sqing wrote: Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Show that $$ \frac{1 - 5a^2}{a + bc} + \frac{1 - 5b^2}{b + ca} + \frac{1 - 5c^2}{c + ab} \ge 3$$ We have \begin{align*} \sum_{cyc}\frac{1-5a^2}{a+bc}&=\sum_{cyc}\frac{(a+b+c)^2-5a^2}{a(a+b+c)+bc}\\ &=\sum_{cyc}\frac{b^2+c^2-4a^2+2(bc+ca+ab)}{(a+b)(a+c)}\\ &=\boxed{\frac{2\sum a^3+\sum bc(b+c)+12abc}{\sum bc(b+c)+2abc}.} \end{align*}Therefore, we need to prove that $$\frac{2\sum a^3+\sum bc(b+c)+12abc}{\sum bc(b+c)+2abc}\ge 3$$$$\iff \boxed{\sum_{cyc}a(a-b)(a-c)\ge 0.}$$Which is just $\textcolor{red}{\textit{Schur's Inequality.}}$ Equality holds iff $a=b=c=\frac 13.$ $\textcolor{green}{\textit{QED.}}$