Given three postive numbers $a, b, c$ so that $a+ b+ c= 3$. Prove that
$$\sum\limits_{sym}\frac{1}{a}\geqq \left (\!\frac{1}{2}+ \frac{5\sqrt{3}}{18}\!\right )(\!a^{2}+ b^{2}+ c^{2}\!)$$
I use discriminant and uvw and find it with the best constant $5\sqrt{3}\div 18$ .
Titu lemma and Holder inequality both are unsuccessful here, so I need to the help. Thanks a real lot
Best Answer
Let $a+b=c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=\frac{u^3v^2}{w^3}-\left(\frac{1}{2}+\frac{5\sqrt3}{18}\right)(3u^2-2v^2).$$ But we see that $f$ decreases, 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 $c=3-2a$, where $0<a<\frac{3}{2}.$
Id est, it's enough to prove that $$\frac{2}{a}+\frac{1}{3-2a}\geq\left(\frac{1}{2}+\frac{5\sqrt3}{18}\right)(2a^2+(3-2a)^2),$$ which is smooth.
I got that the last inequality is equivalent to $$((10+6\sqrt3)a^2-(23+13\sqrt3)a+12+8\sqrt3)(2a-3+\sqrt3)^2\geq0,$$ which is obvious.