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

Question

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!

Answer 1

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 $$a^3+b^3+c^3 \geq ab^2 +bc^2+ca^2.$$

Answer 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.

Answer 3

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.

Answer 4

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 $(2)$.