Prove $3\left(9-5\sqrt{3}\right) \sum \frac{1}{a} \geqslant \sum a^2+\frac32\cdot\frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}$

Question

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

Answer 1

$uvw$ 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.

1. $w^3\rightarrow0^+$.

In this case our inequality is obviously true.

2. 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.