<s>[INST] Prove hard inequality Let $a,b,c>0$, prove that: $$\frac{1}{(2a+b)^2}+\frac{1}{(2b+c)^2}+\frac{1}{(2c+a)^2}\geq\frac{1}{ab+bc+ca}$$
I tried to use the inequality $\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq\frac{(a+b+c)^2}{x+y+z} \forall x,y,z>0$ but that's all I can do. I think this inequality is too tight to use AM-GM or Cauchy Schwarz alone.
 [/INST] After full expanding we need to prove that:
$$\sum_{cyc}(4a^{5}b+4a^{5}c-12a^{4}b^{2}+12a^{4}c^{2}+5a^{3}b^{3}+8a^{4}bc-19a^{3}b^{2}c+5a^{3}c^{2}b-7a^{2}b^{2}c^{2})\geq0$$ or
$$6\sum_{cyc}ab(a^{2}-b^{2}-2ab+2ac)^{2}+
\sum_{sym}(2a^{5}b-a^{3}b^{3}-4a^{4}bc+10a^{3}b^{2}c-7a^{2}b^{2}c^{2})\geq0.$$
Thus, it's enough to prove that:
$$\sum_{cyc}(a^5b+a^5c-a^3b^3-4a^4bc+5a^3b^2c+5a^3c^2b-7a^2b^2c^2)\geq0.$$
Can you end it now?
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