Soluție.
a) 1 3 + 1 3 + 5 3 + 6 3 − 7 3 = 0 {\displaystyle 1^{3}+1^{3}+5^{3}+6^{3}-7^{3}=0} .
b) Din punctul a) putem scrie 1 3 + 1 3 + 5 3 + 6 3 = 7 3 {\displaystyle 1^{3}+1^{3}+5^{3}+6^{3}=7^{3}} sau ( 1 7 ) 3 + ( 1 7 ) 3 + ( 5 7 ) 3 + ( 6 7 ) 3 = 1 {\displaystyle \left({\frac {1}{7}}\right)^{3}+\left({\frac {1}{7}}\right)^{3}+\left({\frac {5}{7}}\right)^{3}+\left({\frac {6}{7}}\right)^{3}=1} .
Acum ( 1 7 ) 2018 < ( 1 7 ) 3 , ( 5 7 ) 2018 < ( 5 7 ) 3 , ( 6 7 ) 2018 < ( 6 7 ) 3 {\displaystyle \left({\frac {1}{7}}\right)^{2018}<\left({\frac {1}{7}}\right)^{3},\left({\frac {5}{7}}\right)^{2018}<\left({\frac {5}{7}}\right)^{3},\left({\frac {6}{7}}\right)^{2018}<\left({\frac {6}{7}}\right)^{3}} , de unde ( 1 7 ) 2018 + ( 1 7 ) 2018 + ( 5 7 ) 2018 + ( 6 7 ) 2018 < ( 1 7 ) 3 + ( 1 7 ) 3 + ( 5 7 ) 3 + ( 6 7 ) 3 . {\displaystyle \left({\frac {1}{7}}\right)^{2018}+\left({\frac {1}{7}}\right)^{2018}+\left({\frac {5}{7}}\right)^{2018}+\left({\frac {6}{7}}\right)^{2018}<\left({\frac {1}{7}}\right)^{3}+\left({\frac {1}{7}}\right)^{3}+\left({\frac {5}{7}}\right)^{3}+\left({\frac {6}{7}}\right)^{3}.}
Obținem astfel ( 1 7 ) 2018 + ( 1 7 ) 2018 + ( 5 7 ) 2018 + ( 6 7 ) 2018 < 1 {\displaystyle \left({\frac {1}{7}}\right)^{2018}+\left({\frac {1}{7}}\right)^{2018}+\left({\frac {5}{7}}\right)^{2018}+\left({\frac {6}{7}}\right)^{2018}<1} sau 1 2018 + 1 2018 + 5 2018 + 6 2018 < 7 2018 {\displaystyle 1^{2018}+1^{2018}+5^{2018}+6^{2018}<7^{2018}} .
Deoarece 5 2018 + 6 2018 < 1 2018 + 1 2018 + 5 2018 + 6 2018 < 7 2018 {\displaystyle 5^{2018}+6^{2018}<1^{2018}+1^{2018}+5^{2018}+6^{2018}<7^{2018}} , inegalitatea este demonstrată.
.
sau 
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, de unde 
sau 
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, inegalitatea este demonstrată.
