prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$

Question

let $a,b,c\ge 0$ and such $a+b+c=3$ show that $$(a^3+1)(b^3+1)(c^3+1)\ge 8$$

My research:It seem use Holder inequality,so $$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$ Use AM-GM $$abc\le\dfrac{(a+b+c)^3}{27}=1$$ what? then I think this method is wrong

Answer 1

$\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2\right)=\sum\limits_{cyc}\left(\ln(a^3+1)-\ln2-\frac{3}{2}(a-1)\right)$.

Let $f(x)=\ln(x^3+1)-\ln2-\frac{3}{2}(x-1)$.

Easy to see that $\min\limits_{[0,2]}f=0$.

Indeed, $f'(x)=\frac{3x^2}{x^3+1}-\frac{3}{2}=\frac{3(1-x)(x^2-x-1)}{2(x^3+1)}$ and since $x_{max}=\frac{1+\sqrt5}{2}$,

it remains to check that $f(2)>0$, which is true.

Thus, for $\{a,b,c\}\subset[0,2]$ our inequality is proven.

Let $a>2$.

Hence, $\prod\limits_{cyc}(a^3+1)\geq(8+1)\cdot1\cdot1>8$.

Done!

Answer 2

Let $f(x,y,z) = (x^3 +1)(y^3 +1)(z^3 +1)$ subject to the constraint $g(x,y,z) = x + y + z - 3 =0$ and $ x,y,z \ge 0$.

Then using lagrange multipliers we have

• $f_x = 3x^2(y^3 +1)(z^3 +1)= g_x= \lambda$
• $f_y = 3y^2(x^3 +1)(z^3 +1)= g_y = \lambda$
• $f_z = 3z^2(x^3 +1)(y^3 +1)=g_z = \lambda$
• $g(x,y,z) = x + y + z - 3 =0$

From the first three equations we see $x^2(y^3 +1) = y^2(x^3 +1) = z^2(y^3 +1)$. So then $x =z$, since $x, z \ge 0$. Then clearly $x=y=z$, so plugging into our formula for $g$, we see $x=y=z=1$, so $(1,1,1)$ is a critical point.

Now, $f(1,1,1) = 8$. To see this is a minimum, we see $(0,0,3)$ lies on our curve, and $f(0,0,3) = 28 \gt f(1,1,1) = 8.$

We conclude that $(a^3 +1)(b^3 +1)(c^3 +1) \ge 8$.

Answer 3

WLOG we assume that $(b-1)(c-1)\ge 0$.

By Carlson inequality we have \begin{aligned} {[(a^3+1)(b^3+1)(c^3+1)]}^{\frac{1}{3}}= & \;\;[(a^3+1)(1+b^3)(1+c^3)]^{\frac{1}{3}} \\ \ge & \;\;a+bc\\ = & \;\;3-b-c+bc\\ = & \;\;2+(b-1)(c-1)\\ \ge & \;\;2. \end{aligned} Done.