<s>[INST] 3-variable symmetric inequality Given $a,b,c>0$ satisfying $a^2+b^2+c^2=3$. Prove that
$$2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+3(a+b+c)\geq 15.$$
I've tried to use the inequality $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{9}{a+b+c}$ as well as AM-HM and the condition $a+b+c \leq 3$ but it still doesn't work.
 [/INST] There is also the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$, and $abc=w^3$.
Thus, by AM-GM $v\geq w$ and the condition gives $3u^2-2v^2=1,$ which gives $u=\sqrt{\frac{1+2v^2}{3}}.$
Thus, $$\sum_{cyc}\frac{1}{a}=\frac{ab+ac+bc}{abc}=\frac{3v^2}{w^3}\geq\frac{3}{v}.$$
Id est, it's enough to prove that
$$\frac{2}{v}+3u\geq5$$ or
$$\frac{2}{v}+\sqrt{3(1+2v^2)}\geq5,$$ which is obvious for $\frac{2}{v}\geq5$ or $5v-2\leq0.$
But for $5v-2>0$ we need to prove that
$$v\sqrt{3(1+2v^2)}\geq5u-2$$ or after squaring of the both sides
$$(v-1)^2(3v^2+6v-2)\geq0,$$
which is true for $5v-2>0$.
Done!
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