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Try this beautiful problem from HANOI 2018 based on Inequalities.

## Inequalities – HANOI 2018

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.

• is 1
• is 2
• is 8
• cannot be determined from the given information

### Key Concepts

Inequalities

Algebra

Number Theory

But try the problem first…

Source

HANOI, 2018

Inequalities (Little Mathematical Library) by Korovkin

## Try with Hints

First hint

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)

Second Hint

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 $T=(a-b)^{2018} + (b-c)^{2018}+ (c-a)^{2018}$

$\leq a-b+b-c+a-c=2(a-c) \leq 2$

Final Step

The equality holds if (a,b,c)=(2,2,1) or (a,b,c)=(2,1,1). Then max T = 2