Strategies to prove inequalities with interval notation How to prove a inequalities with interval notation, for example:


  Find minimum of $a^3+b^3+c^3$ with $a,b,c \in [-1;\infty), a^2+b^2+c^2=9$
Clearly $a, b, c$ cannot be all negative.

If $a, b, c$ are all non-negative, then by Power mean inequality $(\frac{a^3+b^3+c^3}{3})^2 \geq (\frac{a^2+b^2+c^2}{3})^3=27$, so $a^3+b^3+c^3 \geq 9\sqrt{3}$ with equality when $a=b=c=\sqrt{3}$.

If exactly 1 of $a, b, c$ is negative, then WLOG assume that $c$ is negative, so $a, b$ are non-negative. Then $c \geq -1$, so $c^2 \leq 1$ so $a^2+b^2 \geq 8$. By Power mean inequality $(\frac{a^3+b^3}{2})^2 \geq (\frac{a^2+b^2}{2})^3 \geq (\frac{8}{2})^3=64$, so $a^3+b^3+c^3 \geq 2\sqrt{64}+c^3 \geq 16-1=15$, with equality when $a=b=2, c=-1$. (and permutations)

If exactly 2 of $a, b, c$ are negative, then WLOG assume that $b, c$ are negative and $a$ is non-negative. We have $b, c \geq -1$, so $b^2, c^2 \leq 1$, so $a^2 \geq 7$, so $a^3+b^3+c^3 \geq (\sqrt{7})^3-1-1=7\sqrt{7}-2$, with equality when $a=\sqrt{7}, b=c=-1$. (and permutations)

Thus $a^3+b^3+c^3 \geq \min(9\sqrt{3}, 15, 7\sqrt{7}-2)=15$, with equality when $a=b=2, c=-1$ and permutations.
By Holder: if $a,b,c&gt;0$
$$(a^3+b^3+c^3)(a^3+b^3+c^3)(1+1+1)\ge (a^2+b^2+c^2)^3$$

then 
$$a^3+b^3+c^3\ge 9\sqrt{3}$$
and when $a=b=c=\sqrt{3}$

case 2:let $a\ge b\ge 0&gt;c\ge -1$
then $a^2+b^2&gt;8$,then $a&gt;2sqrt{2}$,so $a^3+b^3+c^3&gt;a^3+c^3&gt;a^3-1&gt;\sqrt{512}-1&gt;9\sqrt{3}$

case 3:let $a&gt;0&gt;b&gt;c\ge -1$,and we have $a^2&gt;7$,so $a&gt;\sqrt{7}$,then

$a^3+b^3+c^3&gt;\sqrt{343}-2&gt;9\sqrt{3}$