法一：构造一元偏差函数
\({f}^{\text{'}}\left ( {x} \right )=\frac {1-x} {{e}^{x}}\) ，则 \( f\left(x\right)\) 在 \((-\infty ，1)\) 上递增， \((1，+\infty )\) 上递减
令 \( F\left(x\right)=f(1+x)-f(1-x)(x>0)\)
\( \therefore {F}^{\text{'}}\left(x\right)={f}^{\text{'}}(1+x)+{f}^{\text{'}}(1-x)=\frac{-x}{{e}^{1+x}}+\frac{x}{{e}^{1-x}}=\frac{x({e}^{1+x}-{e}^{1-x})}{{e}^{2}}>0\)
\( \therefore F\left(x\right)>F\left(0\right)=0\) 即 \( f(1+x)>f(1-x)\)
不妨设 \( 0<{x}_{1}<1<{x}_{2}\)
\( \therefore f\left({x}_{1}\right)=f\left({x}_{2}\right)=f(1+({x}_{2}-1\left)\right)>f(1-({x}_{2}-1\left)\right)=f(2-{x}_{2})\)
显然 \( 2-{x}_{2}<1\) ，又 \( f\left(x\right)\) 在 \( (-\infty ,1)\) 上递增，则 \( {x}_{1}>2-{x}_{2}\) ，即 \( {x}_{1}+{x}_{2}>2\)
法二：比值代换
由 \( f\left({x}_{1}\right)=f\left({x}_{2}\right)\) 得 \( {x}_{1}{e}^{-{x}_{1}}={x}_{2}{e}^{-{x}_{2}}\) ，则 \( {e}^{{x}_{2}-{x}_{1}}=\frac{{x}_{2}}{{x}_{1}}\)
不妨设 \( 0<{x}_{1}<1<{x}_{2}\) ，令 \( t=\frac{{x}_{2}}{{x}_{1}}(t>1)\) ，则 \({x}_{2}-{x}_{1}>\ln {t}\)
\({\therefore x}_{1}=\frac {\ln {t}} {t-1}\)， \({x}_{2}=\frac {t\ln {t}} {t-1}\)，则\({\therefore x}_{1}+{x}_{2}=\frac {\left ( {t+1} \right )\ln {t}} {t-1}\)
要证： \( {x}_{1}+{x}_{2}>2\) ，只需证： \(\frac {\left ( {t+1} \right )\ln {t}} {t-1}>2\)， 即证：\(\ln {t}>\frac {2\left ( {t-1} \right )} {t+1}\)
令 \(h\left ( {t} \right )=\ln {t}-\frac {2\left ( {t-1} \right )} {t+1}\) ，则 \( {h}^{\text{'}}\left(t\right)=\frac{1}{t}-\frac{4}{(t+1{)}^{2}}=\frac{(t-1{)}^{2}}{t(t+1{)}^{2}}>0\)
\( \therefore h\left(t\right)>h\left(1\right)=0\) 即证
法三：对数均值不等式
设 \( f\left({x}_{1}\right)=f\left({x}_{2}\right)=t\) ，则 \( \frac{{x}_{1}}{{e}^{{x}_{1}}}=t\) ， \( \frac{{x}_{2}}{{e}^{{x}_{2}}}=t\) ， \( ({x}_{1}\ne {x}_{2})\) 两边取对数
\(\ln {{x}_{1}}-{x}_{1}=\ln {t}\)①
\(\ln {{x}_{2}}-{x}_{2}=\ln {t}\)②
①-②得：\(\frac {{x}_{1}-{x}_{2}} {\ln {{x}_{1}}-\ln {{x}_{2}}}=1\)
根据对数平均值不等式\(\frac {{{x}_{1}+x}_{2}} {2}>\frac {{x}_{1}-{x}_{2}} {\ln {{x}_{1}}-\ln {{x}_{2}}}=1\)
\( \therefore {x}_{1}+{x}_{2}>2\)
