Inequality-5
a + 1 1 + b + 1 1 + c + 1 1 1 − a 1 + b 1 + c 1 1 has a minimum value of n m , where a , b , and c are positive numbers, and coprime positive integers m and n .
What is m + n ?
The answer is 4.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Minimum value occurs at, a = b = c = − 0 . 5
2 + 2 + 2 1 − ( − 2 ) + ( − 2 ) + ( − 2 ) 1 = 6 1 − ( − 6 1 ) = 3 1 = n m
m + n = 1 + 3 = 4
Infact this works for all , a = b = c = R − ( 0 , ∞ ) ,where R is any real number.
Doesn't it occur at all a = b = c ?
Log in to reply
Yeah, It does except for a = b = c = 0 , ∞
Thanks for pointing that out:)
Thank you for sharing your solution.
Since a , b , c are all positive, you can't take a = b = c = − 0 . 5 .
Log in to reply
I am trying to tell that this holds true even for negative integers except 0 and ∞ .
Let, a = b = c = − k ⇒ 3 ( 1 − k ) − 3 − k = 3 ( 1 − k + k ) = 3 1
I'm sorry that I dont know the "official" solution.
Maximum or minimum values of symmetric expressions (generally) always occur when all of the symmetric terms are equal
Putting a = b = c we get the expression as 3 a + 1 − 3 a = 3 1
Answer:- 1+3=4
Thank you for sharing your ideas.
Expanding the inequality yields:
a ( a + 1 ) 1 + b ( b + 1 ) 1 + c ( c + 1 ) 1 ≥ 3 1 ( a 1 + b 1 + c 1 ) ( a + 1 1 + b + 1 1 + c + 1 1 ) ≥ 3 1
This is true by Chebyshev's inequality, ⟹ m = 1 and n = 3 ⟹ m + n = 4