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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$!$

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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.

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