<s>[INST] Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$ If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$
My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get :
$$\sum a^4b+3\sum a^3b^2+6\sum a^2b^2c\ge 14\sum a^3bc$$
I don't know what to do next. I think SOS may be helpful but find it difficult to factorize. Any other methods are also welcome but, if possible, can someone help me continue from here?
 [/INST] Now, we need to prove that:
$$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)+4\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)-6\sum_{cyc}(a^3bc-a^2b^2c)\geq0$$ or
$$\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)+$$
$$+4\sum_{cyc}(c^3a^2+c^3b^2-2c^3bc)-3abc\sum_{cyc}(a^2+b^2-2ab)\geq0$$ or
$$\sum_{cyc}(a-b)^2(ab(a+b)+4c^3-3abc)\geq0,$$ which is true by AM-GM:
$$ab(a+b)+4c^3\geq2\sqrt{a^3b^3}+c^3\geq3\sqrt[3]{a^3b^3c^3}=3abc$$
The following inequality a bit of stronger.

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that:
$$\frac{1}{abc}+48\geq\frac{25}{ab+ac+bc}.$$

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