$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis. 
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$. 
To show that formally one can firstly fix $0<c \le 3$, introduce a variable $0\le\alpha\le1$ and express $a$ and $b$ through $\alpha$ 
$$a= \alpha(3-c)$$ $$b=(1-\alpha)(3-c)$$ 
Then we minimize $(a+2)^3+(b+2)^3+(c+2)^3$ w.r.t $\alpha$ treating $c$ as a parameter. We get $\alpha=\frac{1}{2}(3-c)$. Further we maximize $(a+2)^3+(b+2)^3+(c+2)^3$ one more time w.r.t. $c$.
That is quite tedious. I am wondering whether their is a more elegant way. Probably using idea of norms. Basically we need to show that $\lVert(a,b,c)+(2,2,2)\rVert_3 \ge (81)^{1/3}$ while $\lVert(a,b,c)\rVert_1=3$ and $a,b,c>0$. 

Let $f(x) = x^3 $ is a convex function. Using Jenson we get : 
$$ \frac{f(a+2)+f(b+2)+f(c+2)}{3} \geq f(\frac{a+b+c+6}{3})=f(3)=27$$
