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

## Check the Answer

But try the problem first…

Answer: is 2.

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

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

Google