<s>[INST] Prove $3\left(9-5\sqrt{3}\right) \sum \frac{1}{a} \geqslant \sum a^2+\frac32\cdot\frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}$ Let $a,\,b,\,c$ are positive real numbers satisfy $a+b+c=3.$ Prove that
$$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2 + \frac32 \cdot \frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}. \quad (1)$$
Note. From $(1)$ we get
$$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2.$$
It's was posted here.
My solution is write it as SOS
$$\sum \frac{\left[(9-4\sqrt3)c+ab\right](2c+\sqrt3-3)^2(a-b)^2}{24abc} \geqslant 0.$$
Any comments and solutions are welcome and appreciated
 [/INST] $uvw$ kills this problem!
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\frac{9(9-5\sqrt3)v^2}{w^3}\geq9u^2-6v^2+\frac{3(w^3-3(2-\sqrt3)v^2)^2}{2w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=6(9-5\sqrt3)u^4v^2-2(3u^2-2v^2)uw^3-3(w^3-3(2-\sqrt3)uv^2)^2.$$
But $$f''(w^3)=-6<0,$$ which says that $f$ is a concave function.
We know that the concave function gets a minimal value for an extreme value of $w^3$,
which by $uvw$ happens in the following cases.

*

*$w^3\rightarrow0^+$.

In this case our inequality is obviously true.


*Two variables are equal.

Let $b=a$ and $c=3-2a$, where $0<a<1.5.$
Thus, we obtain an inequality of one variable, which easy to check.
I got that it's enough to prove: $$(a-1)^2\left(a-\frac{3-\sqrt3}{2}\right)^2(6-2\sqrt3-a)\geq0,$$ which is obvious.
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