Given three non-negative numbers $a,\!b,\!c$ so that $a+\!b+\!c\!=\!3$. Prove that $\prod\limits_{cyc}(\!2+ a^{2}\!)+ abc\geqq 28$.
Let $a+ b+ c= 3u= 3, ab+ bc+ ca= \frac{3u}{X}, abc= \frac{u^{3}}{wX}$ so $1\leqq X\leqq w$, so
$$\therefore\frac{(16X^{2}- 24X+ 18)w^{2}- 11X+ 1}{w^{2}X^{2}}= \frac{F(w)}{w^{2}X^{2}}\geqq 0$$
$$\because F(w)= (16X^{2}- 24X+ 18)w^{2}- 11X+ 1\geqq 0$$
By discriminant and sos: $(16X^{2}- 24X+ 18)w^{2}- 11X+ 1\geqq (16X^{2}- 24X+ 18)(w- X)^{2}$ .
Best Answer
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $$\prod_{cyc}(2u^2+a^2)+u^3w^3\geq28u^6$$ or $f(w^3)\geq0,$ where $$f(w^3)=w^6-11u^3w^3+16u^6-24u^4v^2+18u^2v^4.$$ But $$f'(w^3)=2w^3-11u^3\leq0,$$ which says that it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$.
Id est, we need to prove that $$(2+a^2)^2(2+(3-2a)^2)+a^2(3-2a)\geq28$$ or $$(a-1)^2(4a^4-4a^3+15a^2-16a+16)\geq0,$$ which is obvious.
Done!