$E = (a + \dfrac{1}{a})^2 + (b + \dfrac{1}{b})^2 + (c + \dfrac{1}{c})^2$
$E = a^2 + b^2 + c^2 + \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + 6$
○ $(a + b + c)^2 = 1 \le 3(a^2 + b^2 + c^2)$ (Theo Cauchy- Schwarz)
$\leftrightarrow a^2 + b^2 + c^2 \ge \dfrac{1}{3}$ (đẳng thức xảy ra $\leftrightarrow a = b = c = 1/3$)
○ $(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c})^2 \le 3(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2})$ (theo Cauchy - Schwarz)
$(a + b + c)(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}) \ge 3. \sqrt[3]{abc}. 3.\sqrt[3]{\dfrac{1}{abc}} = 9$ (theo Cauchy)
$\leftrightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge \dfrac{9}{a + b + c} = 9$
$ 9^2 \le (\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c})^2 \le 3(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2})$
$\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} \ge 27$ (đẳng thức xảy ra $\leftrightarrow a = b = c = 1/3$)
$\rightarrow E \ge 6 + 27 + \dfrac{1}{3} = \dfrac{100}{3}$
Đẳng thức xảy ra $\leftrightarrow a = b = c = 1/3$