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?

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:
$$\frac{1}{abc}+48\geq\frac{25}{ab+ac+bc}.$$

