<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] Suppose $t = \frac{a+b}{2}$ and $c = \max \{a,b,c\}.$ Let
$$f(a,b,c) = \frac{1}{abc}+36 - \frac{21}{ab+bc+ca}.$$
We have
$$f(a,b,c) -f(t,t,c) = \frac{1}{abc}-\frac{4}{c(a+b)^2}+\frac{84}{(a+b)(a+b+4c)}-\frac{21}{ab+bc+ca}$$
$$ = \frac{(a-b)^2}{a+b}\left(\frac{1}{abc(a+b)}-\frac{21}{(ab+bc+ca)(a+b+4c)}\right) \geqslant 0,$$
because
$$(ab+bc+ca)(a+b+4c)=(a+b+c)(ab+bc+ca)(a+b+4c) $$
$$ \geqslant 9abc (a+b+4c) > 21abc(a+b).$$
Thefore
$$f(a,b,c) \geqslant f(t,t,c) = f\left(\frac{1-c}{2},\frac{1-c}{2},c\right)=\frac{4(3c^2-3c+1)(3c-1)^2}{c(3c+1)(c-1)^2} \geqslant 0.$$
The proof is completed.
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