Prove $8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab)$

Question

Let $a\geq0, b\geq0,c\geq0$. Prove that:

$$8abc(a+b+c)^3\leq27(a^2+bc)(b^2+ca)(c^2+ab).$$

I tried to prove this inequality using only the inequality between the arithmetic mean and the geometric mean. Does anyone succeed? Thanks!

Answer 1

Remark $$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$ means cyclic sum. $$\displaystyle\prod f(a,b,c)=f(a,b,c)\cdot f(b,c,a) \cdot f(c,a,b)$$ means cyclic product.

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denote $\displaystyle f(a,b,c):=27\prod(a^2+bc)-8abc\sum a$

then expand it, we obtain

$$f(a,b,c)=27\sum{a^3b^3}+19\sum{a^4bc}-24\sum{a^3b^2c}-24\sum{a^2b^3c}+6a^2b^2c^2$$

It's easy to get an SOS form.

$$f(a,b,c)=22 \sum{a^3 b (b-c)^2}+5 \sum{a^3 (b-c)^2 c}+17 \sum{a (a-b)^2 b^2 c}+2 a b c\sum{a (a-b) (a-c)}\geq 0 $$

Answer 2

Without full expanding:

If $abc = 0$, clearly the inequality is true.

In the following, WLOG, assume that $a \ge b \ge c > 0$.

We split into two cases:

Case 1: $b^2 \ge ca$

Using Holder, we have \begin{align*} \mathrm{RHS} &\ge 27\left(\sqrt[3]{bc \cdot ca \cdot c^2} + \sqrt[3]{a^2\cdot b^2\cdot ab}\right)^3\\ &= 27\left(c\sqrt[3]{abc} + ab\right)^3. \end{align*} It suffices to prove that $$3(c\sqrt[3]{abc} + ab) \ge 2\sqrt[3]{abc}\,(a + b + c)$$ or $$3ab \ge \sqrt[3]{abc}\,(2a + 2b - c)$$ which is true. See Remark 1 at the end.

$\phantom{2}$

Case 2: $b^2 < ca$

Using Holder, we have \begin{align*} \mathrm{RHS} &\ge 27\left(\sqrt[3]{a^2 \cdot ca \cdot ab} + \sqrt[3]{bc\cdot b^2\cdot c^2}\right)^3\\ &= 27\left(a\sqrt[3]{abc} + bc\right)^3. \end{align*} It suffices to prove that $$3(a\sqrt[3]{abc} + bc) \ge 2\sqrt[3]{abc}(a + b + c)$$ or $$3bc \ge \sqrt[3]{abc}\,(2b + 2c - a)$$ which is true. See Remark 2 at the end.

We are done.

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Remark 1:

Let $f(c) = \sqrt[3]{abc}\,(2a + 2b - c)$. We have $f'(c) = \frac{2ab(a + b - 2c)}{3\sqrt[3]{a^2b^2c^2}} \ge 0$ for all $0 < c \le b^2/a$.

It suffices to prove that $$3ab \ge \sqrt[3]{ab \cdot \frac{b^2}{a}}\,\left(2a + 2b - \frac{b^2}{a}\right)$$ or $$a + b^2/a \ge 2b$$ which is true.

Remark 2:

Let $g(a) = \sqrt[3]{abc}\,(2b + 2c - a)$. We have $g'(a) = - \frac{2bc(2a - b - c)}{3\sqrt[3]{a^2b^2c^2}} \le 0$ for all $a \ge b^2/c$.

It suffices to prove that $$3bc \ge \sqrt[3]{\frac{b^2}{c} \cdot bc}\,\left(2b + 2c - \frac{b^2}{c}\right)$$ or $$c + b^2/c \ge 2b$$ which is true.