设 $a , b , c > 0$，且 $a + b + c = 1$，求 ${\left( {a + \dfrac{1}{a}} \right)^2} + {\left( {b + \dfrac{1}{b}} \right)^2} + {\left( {c + \dfrac{1}{c}} \right)^2}$ 的最小值．
【难度】
【出处】
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【答案】
$\dfrac{100}{3}$
【解析】
考虑到取等条件，有\[\begin{split}\sum\limits_{cyc} {{{\left( {a + \dfrac{1}{a}} \right)}^2}} &= 6 + \sum\limits_{cyc} {\left( {{a^2} + \dfrac{1}{{{a^2}}}} \right)}\\& = 6 + \sum\limits_{cyc} {\left( {{a^2} + \dfrac{1}{{81{a^2}}} + \dfrac{1}{{81{a^2}}} + \cdots + \dfrac{1}{{81{a^2}}}} \right)} \\&\geqslant 6 + 246 \cdot {\left[ {\dfrac{1}{{{{81}^{27}} \cdot {{\left( {abc} \right)}^{16}}}}} \right]^{\frac{1}{{30}}}}\\& \geqslant 6 + 246 \cdot {\left[ {\dfrac{1}{{{3^{108}} \cdot {3^{ - 48}}}}} \right]^{\frac{1}{{30}}}} \\&= \dfrac{{100}}{3},\end{split}\]当且仅当 $a = b = c = \dfrac{1}{3}$ 时取得等号．
由柯西不等式得\[\begin{split}{\left( {a + \dfrac{1}{a}} \right)^2} + {\left( {b + \dfrac{1}{b}} \right)^2} + {\left( {c + \dfrac{1}{c}} \right)^2}&\geqslant \dfrac{{{{\left( {a + b + c + \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)}^2}}}{3}\\& = \dfrac{{{{\left( {1 + \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)}^2}}}{3}\\&\geqslant {\dfrac{{\left( {1 + \dfrac{9}{{a + b + c}}} \right)}}{3}^2} \\&= \dfrac{{100}}{3},\end{split}\]当且仅当 $a = b = c = \dfrac{1}{3}$ 时取得等号．
利用切线法构造辅助不等式 ${\left( {x + \dfrac{1}{x}} \right)^2} \geqslant \dfrac{{100}}{9} - \dfrac{{160}}{3}\left( {x - \dfrac{1}{3}} \right)$ 即得．
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