teps wrote: Prove for all positive real numbers $a,b,c$, such that $a^2+b^2+c^2=1$: \[\frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\ge \frac{\sqrt{3}}{1+\sqrt{3}}.\] $\sum_{cyc}\frac{a^3}{b^2+c}=\sum_{cyc}\frac{a^4}{b^2a+ac}\geq\frac{1}{\sum\limits_{cyc}(a^2c+ab)}$. Hence, we need to prove that $1+\sqrt3\geq\sqrt3\sum\limits_{cyc}(a^2c+ab)$, which is true because $1\geq ab+ac+bc$ and $\left(\sum_{cyc}a^2c\right)^2\leq(a^2+b^2+c^2)(a^2c^2+b^2a^2+c^2b^2)=$ $=a^2c^2+b^2a^2+c^2b^2\leq\frac{(a^2+b^2+c^2)^2}{3}=\frac{1}{3}$.

Prove for all positive real numbers $a,b,c$, such that $bc+ca+ab=1$: \[\frac{a^2}{1+a}+\frac{b^2}{1+b}+\frac{c^2}{1+c}\ge \frac{\sqrt{3}}{1+\sqrt{3}}.\]

sqing wrote: Prove for all positive real numbers $a,b,c$, such that $bc+ca+ab=1$: \[\frac{a^2}{1+a}+\frac{b^2}{1+b}+\frac{c^2}{1+c}\ge \frac{\sqrt{3}}{1+\sqrt{3}}.\] Substitution : $ a = \tan(\frac{\alpha}{2}) , b = \tan(\frac{\beta}{2}), c = \tan(\frac{\gamma}{2})$, $\alpha+\beta+\gamma=\pi$ and $f(x)=\frac{ \tan^2(\frac{x}{2})}{1+\tan(\frac{x}{2})}$ - Convex. So by Jensen ....

$LHS\geq \frac{(a+b+c)^2}{3+\sum a}=\frac{t^2}{3+t}\geq \frac{\sqrt{3}}{1+\sqrt{3}}\Leftrightarrow t=\sum a\geq \sqrt{3}$. it's true because: $(\sum a)^2\geq 3.\sum ab=3\rightarrow \sum a\geq \sqrt{3}$

teps wrote: Prove for all positive real numbers $a,b,c$, such that $a^2+b^2+c^2=1$: \[\frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\ge \frac{\sqrt{3}}{1+\sqrt{3}}.\] $\left(\frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\right)\left((ab^2+ac)+(bc^2+ba)+(ca^2+cb)\right) \ge 1$ by Cauchy.. So, its enough to show, $1+\sqrt{3} \ge \sqrt{3}((ab^2+ac)+(bc^2+ba)+(ca^2+cb))$ But, $ab+bc+ca \leq 1$ So, it is enough to show $\frac{1}{3} \ge (ab^2+bc^2+ca^2)^2$ Homogenising, $6(\sum_{cyclic} bc^3a^2) \le \sum a^6+ 3(\sum_{cyclic} a^4b^2)+6a^2b^2c^2$ which is true by AM GM, by the following, $a^2b^2c^2+a^2b^2c^2+b^6+b^4c^2+b^4c^2+c^4a^2 \ge 6a^2b^3c^2$ and so on..