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