How to prove this inequality using AM-GM?

Question

Suppose $a,b,c$ are positive real numbers. Then prove that $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(\frac{a+b+c}{3}\Big)\Big(abc\Big)^\frac{2}{3}\tag{*}$$

My approach: From AM-GM $$\Big(\frac{a+b}{2}\Big)\ge\Big(ab\Big)^\frac{1}{2}\tag{1}$$ Similarly, $\Big(\frac{b+c}{2}\Big)\ge\Big(bc\Big)^\frac{1}{2}\tag{2}$ and $\Big(\frac{c+a}{2}\Big)\ge\Big(ca\Big)^\frac{1}{2}\tag{3}$ Multiplying $(1), (2)$ and $(3)$, we get $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(abc\Big)\tag{4}$$ Which can be written as $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(abc\Big)^\frac{2}{3}\Big(abc\Big)^\frac{1}{3}\tag{4*}$$ Now, again from AM-GM $$\Big(\frac{a+b+c}{3}\Big)\ge\Big(abc\Big)^\frac{1}{3}\tag{5}$$ But now I am stuck. What should I do to get to $(*)$ ?

Answer 1

The inequality $$ \sum_{sym} a^2 b^1 c^0 \geq \sum_{sym} a^1 b^1 c^1 $$ holds by bunching / Muirhead's inequality and it is equivalent to $$ \frac{a+b}{2}\cdot\frac{a+c}{2}\cdot\frac{b+c}{2}\geq \frac{a+b+c}{3}\cdot\frac{ab+ac+bc}{3}.$$ The given inequality is weaker than the latter, since $\frac{ab+ac+bc}{3}\geq\left(abc\right)^{2/3}$ holds by AM-GM.
As pointed out in the comments, the given inequality is equivalent to $$ R^3 r \geq \tfrac{8}{27}\Delta^2 $$ for triangles.

Answer 2

By AM-GM: $$\prod_{cyc}(a+b)-\frac{8}{9}(a+b+c)(ab+ac+bc)=\frac{1}{9}\sum_{cyc}(a^2b+a^2c-2abc)=$$ $$=\frac{1}{9}\sum_{cyc}(a^2c+b^2c-2abc)\geq\frac{1}{9}\sum_{cyc}(2\sqrt{a^2c\cdot b^2c}-2abc)=0.$$ Id est, by AM-GM again we obtain: $$\prod_{cyc}\left(\frac{a+b}{2}\right)^3=\frac{1}{512}\left(\prod_{cyc}(a+b)\right)^3\geq\frac{1}{512}\cdot\frac{512}{729}(a+b+c)^3(ab+ac+bc)^3\geq$$ $$\geq\frac{1}{27}\left(\frac{a+b+c}{3}\right)^3\left(3\sqrt[3]{a^2b^2a^2}\right)^3=\left(\frac{a+b+c}{3}\right)^3a^2b^2c^2$$ and we are done!