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

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! 
