some idea?

$\iff (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3$ $\iff (a+b+c-1)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1)\ge 4$ $\iff (a+b+c-1)(\frac{(a+b+c)^2-4}{2abc}-\frac{1}{2})\ge 4$ Let $a+b+c=x$. After using the result: $abc\le \frac{(x-2)^2}{4-x}$ we need to prove $ (x-1)(\frac{(x+2)(4-x)}{2(x-2)}-\frac{1}{2})\ge 4 \ \ \iff \ \ \frac{(3-x)(x^2+x+2)}{2(x-2)}\ge 0$ It is true, because $2\le x\le 3$

parmenides51 wrote: Let $a, b, c$ be positive numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$\frac{a + b}{c} +\frac{b + c}{a} +\frac{c + a}{b} \ge a + b + c + \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, and $abc=w^3$. Thus, by the Carnot's theorem $u\leq1$, by Maclaurin $u\geq v\geq w$ and the condition gives: $$4=9u^2-6v^2+w^3\leq9u^2-6v^2+v^3,$$which gives $$u\geq\frac{1}{3}\sqrt{4+6v^2-v^3}.$$Now, since $u\leq1$, it's enough to prove that: $$\sum_{cyc}\frac{a+b-1}{c}\geq3$$and after using Chebyshov it's enough to prove that: $$\sum_{cyc}(a+b-1)\sum_{cyc}\frac{1}{c}\geq9$$or $$(2u-1)v^2\geq w^3,$$for which it's enough to prove that $$(2u-1)v^2\geq v^3$$or $$2u\geq1+v,$$for which it's enough to prove that: $$\frac{2}{3}\sqrt{4+6v^2-v^3}\geq1+v$$or $$(v-1)^2(4v-7)\leq0,$$which is obvious.

parmenides51 wrote: Let $a, b, c$ be positive numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$\frac{a + b}{c} +\frac{b + c}{a} +\frac{c + a}{b} \ge a + b + c + \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$$ $$a^2+b^2+c^2+abc = 4 \iff x+y+z+2=xyz\iff \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$$Where $a=\frac{2}{\sqrt{yz}},b=\frac{2}{\sqrt{zx}},c=\frac{2}{\sqrt{xy}}.$

Attachments: