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

A: 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.
A: I found another SOS proof$:$
After homogenising the inequality become$:$
$$a \left( b-c \right) ^{4}+b \left( c-a\right) ^{4}+c \left( a-b
 \right) ^{4}+3\,{a}^{3} \left( b-c \right) ^{2}+3\,{b}^{3} \left( c-a \right) ^{2}+3\,{c}^{3} \left( a-b \right) ^{2}\geqslant 0$$
Or another SOS$:$ $$\sum  \left( {a}^{2}b+a{b}^{2}-6\,abc+3\,{c}^{2}a+3\,{c}^{2}b+{c}^{3}
 \right)  \left( a-b \right) ^{2}\geqslant 0$$
which is easy to prove$:$ $$ {a}^{2}b+a{b}^{2}-6\,abc+3\,{c}^{2}a+3\,{c}^{2}b+{c}^{3}
  \geqslant 0.$$
Can you end it now?
A: Another way.
Let $a=\min\{a,b,c\},$ $b=a+u$ and $c=a+v$.
Thus, $u\geq0$, $v\geq0$ and
$$\sum_{cyc}(a^4b+a^4c+3a^3b^2+3a^3c^2-14a^3bc+6a^2b^2c^2)=$$
$$=6(u^2-uv+v^2)a^3+9uv(u+v)a^2+$$
$$+2(u^4-2u^3v+12u^2v^2-2uv^3+v^4)a+uv(u+v)^3\geq0.$$
A: I think, the shortest way here it's using $uvw$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=\frac{3u^3}{w^3}+4-\frac{7u^2}{v^2}.$$
Now we see that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.
We could have seen it before because the starting inequality it's a polynomial symmetric inequality of fifth degree.
Now, after homogenization we need to prove that:
$$\frac{(a+b+c)^3}{abc}+36\geq\frac{21(a+b+c)^2}{ab+ac+bc}.$$
Since the last inequality is homogeneous and symmetric, it's enough to assume $b=c=1$, which gives:
$$(a-1)^2(a^2-2a+4)\geq0,$$ which is obvious.
