# For $a,b,c>0,\,a+b+c=3,$ prove $\sum\limits_{cyc}\,\frac{1}{a}\geqq\left(\frac{1}{2}+\frac{5}{18}\,\sqrt{3}\right)(a^2+b^2+c^2)$

Given $$a,\,b,\,c> 0$$ such that$$:$$ $$a+ b+ c= 3.$$ Prove$$:$$ $$\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}\geqq \left ( \frac{1}{2}+ \frac{5}{18}\,\sqrt{3} \right )(\,a^{\,2}+ b^{\,2}+ c^{\,2}\,)$$ I find $$constant= \frac{1}{2}+ \frac{5}{18}\,\sqrt{3}$$ by using my discriminant skills$$,$$ but the equality condition is strange because I tried the same$$:$$ $$\lceil$$ https://math.stackexchange.com/a/2836680/552226 $$\rfloor$$ without success$$!$$

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 $$0
Id est, it's enough to prove that $$\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.