Try this beautiful problem from HANOI 2018 based on Inequalities.
Let a,b,c denote the real numbers such that \(1\leq a,b,c \leq 2\). Consider T= \((a-b)^{2018}+(b-c)^{2018}+(c-a)^{2018}\). Find the largest possible value of T.
Inequalities
Algebra
Number Theory
Answer: is 2.
HANOI, 2018
Inequalities (Little Mathematical Library) by Korovkin
Here without loss of generality, one can assume that \(1 \leq c \leq b \leq a \leq2\) Then \(0 \leq a-b \leq 1\) and \((a-b)^{2018} \leq a-b \) and the equality holds if a=b or (a,b)=(2,1)
by the same way \(0 \leq b-c \leq 1\) then \((b-c)^{2018} \leq b-c\), \(0 \leq a-c \leq 1\) then \((c-a)^{2018} \leq a-c\) Then \(T=(a-b)^{2018} + (b-c)^{2018}+ (c-a)^{2018}\)
\(\leq a-b+b-c+a-c=2(a-c) \leq 2\)
The equality holds if (a,b,c)=(2,2,1) or (a,b,c)=(2,1,1). Then max T = 2
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