<s>[INST] 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!
 [/INST] 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
$$a^3+b^3+c^3 \geq ab^2 +bc^2+ca^2.$$
</s>