<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] It's obviously true by $uvw$, but the Tangent Line method also gives a simple solution:
$$\sum_{cyc}\left(\frac{2}{a}+3a-5\right)=\sum_{cyc}\frac{(a-1)(3a-2)}{a}=$$
$$=\sum_{cyc}\left(\frac{(a-1)(3a-2)}{a}-\frac{a^2-1}{2}\right)=\sum_{cyc}\frac{(a-1)^2(4-a)}{2a}\geq0.$$
The last inequality is true because $3=a^2+b^2+c^2>a^2,$ which gives $0<a<\sqrt3.$
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