$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$ Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$
This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality with brutal computation method and assume $x = \min \{ x,y,z \}$, $y = x+a$, $z = x + b$, etc. I wonder if we can prove it using classical inequality.
 A: By Cauchy-Schwarz Inequality we have
$\begin{align}
~~~~~~\sum_\text{cyc}\left(a+3b\right)\sum_\text{cyc}\dfrac{a^{2}+3b^{2}}{a+3b}\ge\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\cdot
\end{align}$
So it suffices to show that
$\begin{align}
~~~~~~&\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\ge3\sum_\text{cyc}\left(a+3b\right)\iff\\&\iff\sqrt{a^{2}+3b^{2}}+\sqrt{b^{2}+3c^{2}}+\sqrt{c^{2}+3a^{2}}\ge\sqrt{12\left(a+b+c\right)}\cdot
\end{align}$
With $a^{2}+b^{2}+c^{2}=3$, the last inequality is a "known unsolved inequality" (and it's true).
See:
$\sum\limits_{cyc} \sqrt{a^2+3b^2}\geq \sqrt{12(a+b+c)}$
Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$
a^2+b^2+c^2=3
Inequality 88
A: If you are unable to solve this equation by other means, this may help
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\ge 3 
\implies(a,b,c)\\= (-3,3,3), (0,1,1), (1,0,1),(1,1,0), (1,1,1), (3,-3,3), (3,3,-3) $$
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}= 4\implies 
(a,b,c)\\
=(-5,3,7), (-4,4,4), (-3,9,5),(-2,2,0), (-2,4,2), (0,-2,2), (2,-2,4), (2,0,-1) $$
WolframAlpha has a solution
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}=  5\implies 
(a,b,c)\\
= (-5,5,5), (5,-5,5), (5,5,5) $$
You can continue this for $6,7,8,\cdots$ and get different solutions for each. This may not be the "proper" way to solve an inequality but it may help in revealing "patterns".
