a^2+b^2+c^2=3

[Marius Stănean]

$a,b,c\ge0,\; a^2+b^2+c^2=3\Longrightarrow$ $\dfrac{1}{1+b^2c^2}+\dfrac{1}{1+c^2a^2}+\dfrac{1}{1+a^2b^2}+\dfrac 12\ge \dfrac{6}{a+b+c}$