Fie a,b numere reale. Demonstrati ca min{a,b}=(1/2)*(a+b-|a-b|).
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REZOLVARE
Daca a<=b:
min{a,b}=a si (1/2)*(a+b-|a-b|)=(1/2)*(a+b-(b-a))=(1/2)*(2a)=a=min{a,b}.
Daca a>b:
min{a,b}=b si (1/2)*(a+b-|a-b|)=(1/2)*(a+b-(a-b))=(1/2)*(2b)=b=min{a,b}.
Din cele doua cazuri discutate rezulta concluzia.