if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$

Question

Let $a,b,c>0,abc=1$ show that $$a^3+b^3+c^3+\dfrac{256}{(a+1)(b+1)(c+1)}\ge 35\tag{1}$$ iff $a=b=c=1$

I know use AM-GM inequality $$a^3+b^3+c^3\ge 3abc=3$$ and $$(a+1)(b+1)(c+1)\ge 2\sqrt{a}\cdot 2\sqrt{b}\cdot 2\sqrt{c}=8$$

In this way, will lead to inequality reverse, but $(1)$ seem is right,so How to prove this inequality

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $f(u)\geq0$, where $$f(u)=27u^3-27uv^2+\frac{256w^5}{2w^2+3uw+3v^2}-32w^3$$ But by AM-GM $f'(u)=81u^2-27v^2-\frac{768w^6}{(2w^2+3uw+3v^2)^2}\geq81u^2-27v^2-12w^2>0$,

which says that $f$ is an increasing function.

Thus, it's enough to prove our inequality for a minimal value of $u$.

$a$, $b$ and $c$ are positive roots of the equation $x^3-3ux^2+3v^2x-w^3=0$ or

$3ux^2=x^3+3v^2x-w^3$, which says that

the parabola $y=3ux^2$ and the graph of $y=x^3+3v^2x-w^3$ have three common points.

Easy to see (draw it!) that $u$ gets a minimal value, when the parabola touches to the graph of $y=x^3+3v^2x-w^3$, which happens for equality case of two variables.

Id est, it remains to prove our inequality for $b=a$ and $c=\frac{1}{a^2}$, which gives

$(a-1)^2(2a^{11}+8a^{10}+18a^9-3a^8-92a^7+5a^6+32a^5+24a^4+16a^3+9a^2+4a+1)\geq0$,

which is true and smooth.

Done!

Answer 2

Alternative proof

WLOG, assume $c \in (0, 1]$.

Denote the inequality by $f(a, b, c) \ge 0$. We have \begin{align} &f(a, b, c) - f(\sqrt{ab}, \sqrt{ab}, c)\\ =\ & (a + b - 2\sqrt{ab})\Big[(a+b)^2 + (a+b)\cdot 2\sqrt{ab} + (2\sqrt{ab})^2 - 3ab\Big]\\ &\qquad - \frac{256(a + b - 2\sqrt{ab})}{(c+1)(ab + a+b+1)(ab + 2\sqrt{ab} + 1)}\\ =\ & (a + b - 2\sqrt{ab}) \\ & \times \left( (a+b)^2 + 2(a+b)\sqrt{ab} + ab - \frac{256}{(c+1)(ab + a+b+1)(ab + 2\sqrt{ab} + 1)}\right)\\ \ge\ & (a + b - 2\sqrt{ab}) \\ & \times \Big( (2\sqrt{ab})^2 + 2(2\sqrt{ab})\sqrt{ab} + ab - \frac{256}{(\frac{1}{ab}+1) (ab + 2\sqrt{ab} +1)(ab + 2\sqrt{ab} + 1)}\Big)\\ =\ & (a + b - 2\sqrt{ab})\Big(9ab - \frac{256ab}{(1+ab)(\sqrt{ab} + 1)^4}\Big)\\ \ge\ & (a + b - 2\sqrt{ab})\Big(9ab - \frac{256ab}{(1+1)(1 + 1)^4}\Big)\\ \ge\ & 0 \end{align} where we have used $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$, $a + b \ge 2\sqrt{ab}$, and $ab = \frac{1}{c} \ge 1$.

Thus, it suffices to prove that $f(\sqrt{ab}, \sqrt{ab}, c) \ge 0$ or $$2(ab)^{3/2} + c^3 + \frac{256}{(\sqrt{ab} + 1)^2(c+1)} - 35 \ge 0$$ or $$2(1/c)^{3/2} + c^3 + \frac{256}{(\sqrt{1/c} + 1)^2(c+1)} - 35 \ge 0.$$ Let $c = u^2$. It suffices to prove that $$\frac{2}{u^3} + u^6 + \frac{256}{(1/u + 1)^2(u^2 + 1)} - 35 \ge 0$$ or \begin{align} &\frac{(u-1)^2}{u^3(u+1)^2(u^2+1)}\\ &\cdot (u^{11}+4 u^{10}+9 u^9+16 u^8+24 u^7+32 u^6+5 u^5-92 u^4-3 u^3+18 u^2+8 u+2)\ge 0 \end{align} which is true. We are done.