parmenides51 wrote: Let $a, b, c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{1}{1+2ab}+\frac{1}{1+2bc}+\frac{1}{1+2ca}\ge 1$$ https://artofproblemsolving.com/community/c6h422566p4655106 Let $a, b, c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{2-c}{1+2ab}+\frac{2-a}{1+2bc}+\frac{2-b}{1+2ca}\ge 1$$ $$\left(\frac{4-a}{1+2ab}\right)a^2+\left(\frac{4-b}{1+2bc}\right)b^2+\left(\frac{4-c}{1+2ca}\right)c^2\ge 3$$ Let $a, b, c > 0$ such that $a^2 + b^2 + c^2 = 3$. Show that $$\frac{1}{a^2 + 2bc} + \frac{1}{b^2 + 2ca} + \frac{1}{c^2 + 2ab} \geq 1$$ Positive reals $a,b,c$ satisfy $a^2+b^2+c^2=3$. Prove: $$\frac{1}{a^3+2bc}+\frac{1}{b^3+2ca}+\frac{1}{c^3+2ab}\geq 1$$

Attachments: