\[F(\,k\,)= (\,16\,X^{\,2}- 24\,X+ 18\,)k^{\,2}- 11\,kX+ 1\geqq 0\]

Given that $ 1\leqq X\leqq k$ $ ,$ prove$ :$ $ $ F(\,k\,)= (\,16\,X^{\,2}- 24\,X+ 18\,)k^{\,2}- 11\,kX+ 1\geqq 0$ $ Original problem$ :$ Given that $ a,\,b,\,c$ are $ 3$ non‚ąínegatve numbers and $ a+ b+ c= 3$ $ ,$ prove$ :$ $ $ (\,2+ a^{\,2}\,)(\,2+ b^{\,2}\,)(\,2+ c^{\,2}\,)+ abc\geqq 28$ $ Let$ :$ $ $ \begin{equation}\begin{split} ab+ bc+ ca & = & \frac{3}{X} \ abc & = & \frac{1}{kX} \end{split}\end{equation}$ $ Back the inequality$ :$ $ $ F(\,X\,)\geqq 0$ $ So I tried$ :$ $ $ F(\,k\,)= F(\,X\,)+ F(\,k- X\,)$ $ But I can$ ‘$ t continue$ :$ $ $ F(\,k- X\,)= (\,k- X\,)(\,16\,X^{\,3}+ 16\,X^{\,2}k- 24\,X^{\,2}- 24\,kX+ 7\,X+ 18\,k\,)\geqq 0$ $ Much harder$ !$ I need to the help$ !$ Thanks$ !$