How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:

Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$

I used Muirhead's inequality to do the question (you can try out yourself):

 By Muirhead's inequality, $$\begin{align}7(ab+bc+ca)&=7(a+b+c)(ab+bc+ca)\\&=21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^2b\\&\le21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^3\\&=2(a+b+c)^3+9abc\\&=2+9abc\end{align}$$As $(3,0,0)$ majorizes $(2,1,0)$. 

Is above proof correct? Also, can we find a proof not using Muirhead's inequality?
Any help is appreciated!

A proof by SOS:
$$2+9abc-7(ab+ac+bc)=2(a+b+c)^3+9abc-7(a+b+c)(ab+ac+bc)=$$
$$=\sum_{cyc}(2a^3+6a^2b+6a^2c+4abc+3abc-7a^2b-7a^2c-7abc)=$$
$$=\sum_{cyc}(a^3-a^2b-ab^2+b^3)=\sum_{cyc}(a-b)^2(a+b)\geq0.$$
Also, $uvw$ kills it immediately. 
