<s>[INST] Prove $\sum\limits_{cyc} \frac{(a+2b+3c)^2}{(2a+b)^2} \geqslant 5 + 7 \cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$ for positive $a,b,c$ For $a,b,c >0$, prove
$$\sum_{cyc} \frac{(a+2b+3c)^2}{(2a+b)^2} \geqslant 5 + 7 \cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$$
My notation
$$\sum_{cyc}a^2b= a^2b+b^2c+c^2a$$
What I try:
1. Using C-S inequality
$$\sum_{cyc} \frac{(a+2b+3c)^2}{(2a+b)^2} \geqslant \frac{\left(6a+6b+6c \right)^2}{\sum\limits_{cyc}(2a+b)^2}$$
It turns out that
$$\frac{\left(6a+6b+6c \right)^2}{\sum\limits_{cyc}(2a+b)^2} \leqslant 5 + 7 \cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$$ 


*I try to change the variables $x=2a+b,y =2b+c, z=2c+a.$ But the new inequality is not better than the original one due to the term $ab+bc+ca$ in the denominator.

 [/INST] Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$. 
Hence, 
$$(ab+ac+bc)\prod\limits_{cyc}(2a+b)^2\left(\sum\limits_{cyc} \frac{(a+2b+3c)^2}{(2a+b)^2}-5-\frac{7(a^2+b^2+c^2)}{ab+bc+ca}\right)=$$
$$=5103(u^2-uv+v^2)a^6+1458(7u^3-5u^2v+5uv^2+7v^3)a^5+$$
$$+81(103u^4-41u^3v-6u^2v^2+259uv^3+103v^4)a^4+$$
$$+324(11u^5+u^4v-18u^3v^2+40u^2v^3+47uv^4+11u^5)a^3+$$
$$+9(88u^6+102u^5v-390u^4v^2+227u^3v^3+954u^2v^4+558uv^5+88v^6)a^2+$$
$$+36(2u^7+9u^6v-21u^5v^2-11u^4v^3+46u^3v^4+57u^2v^5+21uv^6+2v^7)a+$$
$$+(36u^6-52u^5v-87u^4v^2+62u^3v^3+237u^2v^4+164uv^5+36v^6)uv\geq0$$
Done!
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