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

This can be done automagically using Lagrange multipliers (like most inequalities of this kind...)
Take $f=\frac{a}{b+c+3}+\frac{c}{a+b+3}+\frac{b}{a+c+3}$, $g=a^2+b^2+c^2$, and find the critical points of $f$ subject to the condition that $g=3$. In Mathematica, this can be done as follows:
In[15]:= f = a/(b + c + 3) + b/(a + c + 3) + c/(a + b + 3);

In[16]:= g = a^2 + b^2 + c^2;

In[17]:= Solve[Flatten[{g == 3, 
   Table[D[f + q g, v] == 0, {v, {a, b, c}}]}], 
   {a, b, c, q}, Reals]

Out[17]= {{a -> -1, b -> -1, c -> -1, q -> 3/2}, {a -> 1, b -> 1, c -> 1, q -> -(3/50)}}

This shows that there is one critical point (with positive coordinates) at $(a,b,c)=(1,1,1)$.
All we have left is to check that at this point we have a minimum (or a maximum... :P )
