Is this possible to find solutions?
Real numbers a , b , and c are such that a ≤ b ≤ c and ( 1 + a 1 ) ( 1 + b 1 ) ( 1 + c 1 ) = 2
Are there any solutions satisfying these conditions?
Bonus: Prove the correct answer.
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3 solutions
It is easy to check the a = b = c case. In this case, the given equation becomes ( 1 + a 1 ) 3 = 2 , or, as a real solution, 1 + a 1 = 3 2 , or a = b = c = 3 2 − 1 1 . Hence the answer is yes
Perhaps I'm missing something but a = b = c = 3 2 − 1 1 seems to work, so the answer is yes .
Unsolicited L A T E X advice:
You can type the radical like this:
\sqrt[3]{2}
Instead of
^3 \sqrt{2}
Take care!
Let a = − 2 and b = c = 1 . Then a ≤ b ≤ c and ( 1 − 2 1 ) ( 1 + 1 1 ) ( 1 + 1 1 ) = 2 .