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

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−negatve numbers and $$a+ b+ c= 3 ,$$ prove$$:$$ $$(\,2+ a^{\,2}\,)(\,2+ b^{\,2}\,)(\,2+ c^{\,2}\,)+ abc\geqq 28$$ Let$$:$$ $$$$\begin{split} ab+ bc+ ca & = & \frac{3}{X} \ abc & = & \frac{1}{kX} \end{split}$$$$ Back the inequality$$:$$ $$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$$ Much harder$$!$$ I need to the help$$!$$ Thanks$$!$$