Upper bound = maximum

Algebra Level 5

a 1 + b c + b 1 + a c + c 1 + a b \large \dfrac{a}{1+bc} + \dfrac{b}{1+ac} +\dfrac{c}{1+ ab}

If a a , b b and c c are real numbers in the interval [ 0 , 1 ] [0,1] , what is the maximum value of the expression above to 2 decimal places?


The answer is 2.00.

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.

1 solution

Hana Wehbi
Mar 29, 2017

Since a , b , c a, b, c are in the interval [ 0 , 1 ] [0,1] then the supremum or the least upper bound is a number s s where every element in the interval [ 0 , 1 ] [0,1] should be less than s s . Thus, the upper bound is 2 2

If we consider S S as the perimeter of a triangle, then S = a + b + c S= a+b+c .

We know that the sum of two sides is greater than half the perimeter which is S 2 \frac{S}{2}\implies

2 ( a + b ) > S 2(a+b)>S

2 ( c + a ) > S 2(c+a)>S

2 ( c + b ) > S 2(c+b)>S

\implies a b + c + b a + c + c a + b + 2 ( a + b + c ) ( a + b + c ) = 2 \large\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} +\frac{2(a+b+c)}{(a+b+c)}=2

You didn't prove that the maximum value (of 2) is achievable.

Pi Han Goh - 4 years, 2 months ago

Log in to reply

Yeah, I know. I just realized that later. I will post it soon.

Hana Wehbi - 4 years, 2 months ago

a=0, b=1, c=1 and their permutations, serve to show that the maximum of 2 is achievable.

Bob Kadylo - 4 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...