An algebra problem by Aryan Gupta

Please realize that this can be rewritten as $(a-\frac1a)^2+(b-\frac1b)^2+(c-\frac1c)^2=0$

$\boxed{1}$ It has to be $a, b, c \neq 0$ , then applying AM- GM inequality $\displaystyle \frac{ a^2 + b^2 + c^2 + a^{-2} + b^{-2} + c^{-2}}{6} \ge (a^2 \cdot b^2 \cdot c^2 \cdot a^{-2} \cdot b^{-2} \cdot c^{-2})^\frac{1}{6} = 1 \Rightarrow$ the equality is only possible if and only if $a^2 = b^2 = c^2 = a^{-2} = b^{-2} = c^{-2} = 1$ and this is only possible if and only if $a, b, c = \pm 1$ . Hence, the minimum of $a^3 + b^3 + c^3 = - 3$ .

$\boxed{2}$ Let $f$ be the function $f : \mathbb{R} - \{0\} \longrightarrow \mathbb{R}$ such that $f(x) = x^2 + \frac{1}{x^2}$ . This function reaches the minimum as $x = \pm 1$ making the first derivative and seeing that $f^{'} (\pm 1) = 0$ and its second derivative at these points is greater than 0. Thereforeore, $f(x) \ge f(\pm 1) = 2$ and $a^2 + b^2 + c^2 + a^{-2} + b^{-2} + c^{-2} = 6 \iff a^2 = b^2 = c^2 = a^{-2} = b^{-2} = c^{-2} = 1$ . Hence, the minimum of $a^3 + b^3 + c^3 = - 3$ .

$\boxed{3}$ Applying method of Lagrange multipliers. Exercise for the reader.

$\boxed{4}$ $(x - \frac{1}{x})^2 \ge 0 \iff x^2 - 2 + \frac{1}{x^2} \ge 0 \iff x^2 + \frac{1}{x^2} \ge 2...$