[不等式] can's old bat dang thuc

Prove that for any real numbers $ a,b,c$ such that $ a+b+c=3$, then
\[ \frac{1}{2a^2+7}+\frac{1}{2b^2+7}+\frac{1}{2c^2+7} \le \frac{1}{3}.\]
Equality holds for $ (a,b,c)=1,1,1)$ or $ (a,b,c)=\left( 2,\frac{1}{2},\frac{1}{2}\right).$