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: 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
$$a^3+b^3+c^3 \geq ab^2 +bc^2+ca^2.$$
A: 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.
A: Schur's inequality:
$$a^3+b^3+c^3+3abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b) \iff \\
(a+b+c)^3+9abc\ge 4(a+b+c)(ab+bc+ca) \Rightarrow \\
1+9abc\ge 4(ab+bc+ca) \quad (1)$$
Rearrangement:
$$a+b+c=1 \Rightarrow a^2+b^2+c^2=1-2(ab+bc+ca)\ge ab+bc+ca \Rightarrow \\
1\ge 3(ab+bc+ca) \quad (2)$$
Now add $(1)$ and $(2)$.
A: 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. 
