C-S: $\sum\frac{a^2}{a^2+abc}\ge\frac{\left(\sum a\right)^2}{\sum a^2+3abc}=\frac{\left(\sum a\right)^2}{\sum a^2+3\sum ab}\ge\frac{\left(\sum a\right)^2}{\left(\sum a\right)^2+\frac{1}{3}\left(\sum a\right)^2}$.

For all positive numbers $a, b, c$ satisfying $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$, prove that: $$ \frac{a^3}{a+bc} + \frac{b^3}{b+ca} + \frac{c^3}{c+ab} \geq \frac{1}{4} (a^2+b^2+c^2).$$$$\sum \frac{a^3}{a+bc} =\sum \frac{a^4}{a^2+abc}\ge\frac{\left(\sum a^2\right)^2}{\sum a^2+3abc}=\frac{\left(\sum a^2\right)^2}{\sum a^2+3\sum ab}\ge\frac{\left(\sum a^2\right)^2}{\sum a^2+3\sum a^2}=\frac{1}{4} (a^2+b^2+c^2)$$

For all positive numbers $a, b, c$ satisfying $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$, prove or disprove $$ \frac{a^2}{a+bc} + \frac{b^2}{b+ca} + \frac{c^2}{c+ab} \geq \frac{1}{4} (a+b+c)$$$$ \frac{a^n}{a+bc} + \frac{b^n}{b+ca} + \frac{c^n}{c+ab} \geq \frac{1}{4} (a^{n-1}+b^{n-1}+c^{n-1})$$Where $n\in N^+.$

sqing wrote: For all positive numbers $a, b, c$ satisfying $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$, prove or disprove $$ \frac{a^2}{a+bc} + \frac{b^2}{b+ca} + \frac{c^2}{c+ab} \geq \frac{1}{4} (a+b+c)$$ Hölder: $\sum\frac{a^3}{a^2+abc}\ge\frac{1}{3}\cdot\frac{\left(\sum a\right)^3}{\sum a^2+3abc}=\frac{1}{3}\cdot\frac{\left(\sum a\right)^3}{\sum a^2+3\sum ab}\ge\frac{1}{3}\cdot\frac{\left(\sum a\right)^3}{\left(\sum a\right)^2+\frac{1}{3}\left(\sum a\right)^2}$. sqing wrote: For all positive numbers $a, b, c$ satisfying $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$, prove or disprove $$ \frac{a^n}{a+bc} + \frac{b^n}{b+ca} + \frac{c^n}{c+ab} \geq \frac{1}{4} (a^{n-1}+b^{n-1}+c^{n-1})$$Where $n\in N^+.$ For $n=1$ see #2. If $n\in\mathbb{Z}_{\ge 2}$ then homogenizing and using AM-GM we obtain $$\sum\frac{a^{n+1}}{a^2+abc}=\sum\frac{a^{n+1}}{a^2+\sum ab}=\sum\frac{a^{n+1}}{(a+b)(a+c)}\ge\frac{4a^{n+1}}{(2a+b+c)^2}$$WLOG $a+b+c=3$, then it suffices to prove $\sum\frac{4a^{n+1}}{(3+a)^2}\ge\frac{1}{4}\sum a^{n-1}$ or that $f(x)\ge 0$ for $0\le x\le 3$, where $$f(x)=\frac{4x^{n+1}}{(3+x)^2}-\frac{x^{n-1}}{4}+\frac{3(1-x)}{8}$$Equivalently $f(x)=\frac{3}{8(3+x)^2}\cdot g(x)=\frac{3}{8(3+x)^2}(x-1)\left(10x^{n-1}\left(x+\frac{3}{5}\right)-(x+3)^2\right)$. If $3\ge x\ge 1$ then $x^{n-1}\ge 1$ and $g(x)\ge (x-1)\left(10\left(x+\frac{3}{5}\right)-(x+3)^2\right)=(x-1)^2(3-x)\ge 0$. If $1\ge x\ge 0$ then $x\ge x^{n-1}$ and $g(x)\ge (x-1)\left(10x\left(x+\frac{3}{5}\right)-(x+3)^2\right)=9(x-1)^2(x+1)\ge 0$. Hence $$\sum\frac{4a^{n+1}}{(3+a)^2}\ge\frac{1}{4}\sum a^{n-1}+\frac{3}{8}\sum (a-1)=\frac{1}{4}\sum a^{n-1}$$as desired.