$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$

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


Because $(a+2)^3+(b+2)^3+(c+2)^3\geq\frac{1}{9}(a+2+b+2+c+2)^3=81$.


Alternately, here's a hint $$(x+2)^3-27 -27(x-1) =(x-1)^2(x+8)\geqslant 0$$

Tags:

Optimization

Inequality

Real Analysis

Jensen Inequality

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