<s>[INST] If $a,b,c > 0$ satisfy $a^2+b^2+c^2=3$ then $\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}$ 
Given $a,b,c > 0$ satisfying the condition $$a^2+b^2+c^2=3,$$
  prove that
  $$\frac{a}{b+c+3}+\frac{b}{a+c+3}+\frac{c}{a+b+3} \geq \frac{3}{5}.$$

Thank you all
 [/INST] The inequality is not TRUE.  For example. $a=\sqrt{2}$, b=1, c=0, then
$\frac{\sqrt{2}}{4} + \frac{1}{3+\sqrt{2}} + 0 = 0.3536 + 0.2265 < 0.6$
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