INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

March 17, 2020

Problem based on Inequalities | HANOI 2018

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



Number Theory

Check the Answer

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

Subscribe to Cheenta at Youtube

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.