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!

Your proof looks good.

Here's an unsophisticated alternative proof, using only elementary algebra . . .

We don't even need $a,b,c$ to be nonnegative.

As shown below, if $a,b,c\;$are real numbers such that $a+b+c=1$, and if at least one of $a,b,c\;$is between $-1$ and ${\large{\frac{7}{9}}}$ inclusive, then the inequality holds.

Without loss of generality, assume $-1\le a\le {\large{\frac{7}{9}}}$.

Replacing $c\;$by $1-a-b$, we get
\begin{align*}
&(2+9abc)-7(ab+bc+ca)\\[4pt]
=\;&\bigl(2+9ab(1-a-b)\bigr)-7\bigl(ab+(1-a-b)(a+b)\bigr)\\[4pt]
=\;&(7-9a)b^2+\bigl((7-9a)(a-1)\bigr)b+(2+7a^2-7a)\\[4pt]
\end{align*}
which is nonnegative since:


*If $a={\large{\frac{7}{9}}}$, it evaluates to ${\large{\frac{64}{81}}}$.$\\[10pt]$

*If $-1\le a < {\large{\frac{7}{9}}}$, then regarded as a quadratic in the variable $b$, its leading coefficient is positive, and its discriminant
$$
\bigl((7-9a)(a-1)\bigr)^2-4(7-9a)(2+7a^2-7a)
$$
factors as
$$
(a+1)(1-3a)^2(9a-7)
$$
which is nonpositive.
