3-variable symmetric inequality

Question

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.

Answer 1

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 $0<a<\sqrt3.$

Answer 2

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!