How can we not use Muirhead's Inequality for proving the following inequality?

Yes, your proof is correct. I have a similar proof where, for the last step, we need just the AM-GM inequality.

So it suffice to show that $$2(a+b+c)^3 + 9abc \geq 7(a+b+c)(ab+bc+ca)$$ that is $$2(a^3+b^3+c^3) \geq (a^2b +b^2c+c^2a)+ (ab^2 +bc^2+ca^2).$$ Now the inequality $$a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a$$ follows from AM-GM inequality $$x + y + z \geq 3 \sqrt[3]{xyz}.$$ By letting $x = y = a^3$ and $z = b^3$ we get $$a^3 + a^3 + b^3 \geq 3 \sqrt[3]{a^3a^3b^3} = 3a^2b.$$ Similarly, we find \begin{align*} b^3 + b^3 + c^3 &\geq 3 \sqrt[3]{b^3b^3c^3} = 3b^2c \\ c^3 + c^3 + a^3 &\geq 3 \sqrt[3]{c^3c^3a^3} = 3c^2a \end{align*} Add all together and we are done. By symmetry also the other one holds $$a^3+b^3+c^3 \geq ab^2 +bc^2+ca^2.$$

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.