By AM-GM $$\sum_{cyc}\frac{a^3+a^2}{1+bc}=\sum_{cyc}\frac{a^3+a^2}{(a+1)bc}=\sum_{cyc}\frac{a^2}{bc}\geq 3$$

Postive real numbers $a, b, c$ satisfy $a+b+c=3$. Show that$$\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}\geq3 $$Postive real numbers $a, b, c$ satisfy $\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}=3$. Show that $$a+b+c\leq3 $$

augustin_p wrote: Postive real numbers $a, b, c$ satisfy $abc=1$. Show that $$\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}\geq3.$$ $\frac{a^3+a^2}{1+bc}=a^3$, then its obvious.

Let $a, b, c>0 $ and $a+b+c=3$. Show that $$\frac{a^3+a^2}{1+2bc}+\frac{b^3+b^2}{1+2ca}+\frac{c^3+c^2}{1+2ab} \geq 2$$$$ \frac{a^3+2a^2}{1+bc}+\frac{b^3+2b^2}{1+ca}+\frac{c^3+2c^2}{1+ab}\geq\frac{9}{2}$$

Thanks.

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sqing wrote: Postive real numbers $a, b, c$ satisfy $a+b+c=3$. Show that$$\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}\geq3 $$Postive real numbers $a, b, c$ satisfy $\frac{a^3+a^2}{1+bc}+\frac{b^3+b^2}{1+ca}+\frac{c^3+c^2}{1+ab}=3$. Show that $$a+b+c\leq3 $$

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$$\sum_{cyc}\frac{a^3+a^2}{1+bc}\overset{\text{Generalized T2'S}}{\ge}\frac{\sum_{cyc}a^3+\sum_{cyc}a^2}{3^{-1}\left(3+\sum_{cyc}ab\right)}=\frac{3\left(\sum_{cyc}a^3+\sum_{cyc}a^2\right)}{3+\sum_{cyc}a^2}$$$$\text{ Therefore the inequality transforms into }$$$$\frac{3\left(\sum_{cyc}a^3+\sum_{cyc}a^2\right)}{3+\sum_{cyc}a^2}\ge3\Longleftrightarrow\frac{\sum_{cyc}a^3+\sum_{cyc}a^2}{3+\sum_{cyc}ab}\ge1\Longrightarrow\sum_{cyc}a^3+\sum_{cyc}a^2\ge3+\sum_{cyc}ab$$$$\text{ which is clearly true as }$$$$\sum_{cyc}a^3\overset{\text{AM-GM}}{\ge}3\sqrt[3]{a^3b^3c^3}=3abc=3\text{ and }\sum_{cyc}a^2\ge\sum_{cyc}ab$$$$\blacksquare.$$