# A question on finding maximum

IF a+b+c=1,then find maximum value of ab+bc+ca? (a,b,c are positive numbers)

Note by Shubham Bagrecha
5 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

$(a+b+c)^{2}=1$ $a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=1$ $a^{2}+b^{2}+c^{2}-ab-bc-ca+3ab+3bc+3ca=1$ $\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]+3(ab+bc+ca)=1$ $3(ab+bc+ca)$ will be maximum when $\frac {1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]=0$ i.e. is minimum =0. Therefore max. $ab+bc+ca=\frac {1}{3}$

- 5 years, 1 month ago

Nice proof!

- 5 years ago

AM-GM kills it

square a+b+c to get a^2 + b^2 + c^2 + 2(ab+bc+ca) = 1. Note that a^2 + b^2 + c^2 >= ab + bc + ca by AM-GM. thus ab + bc + ca <= 1/3 with equality iff a = b =c = 1/3

- 5 years, 1 month ago

The result also follows from the Rearrangement-Inequality

- 5 years, 1 month ago

solution?

- 5 years, 1 month ago

You can also use LaGrange Multipliers to solve for the maximum value right?

- 5 years ago

1/3

- 5 years, 1 month ago

1/3 is the maximum

- 5 years, 1 month ago

0.33

- 5 years, 1 month ago

0.333333... To be precise (or simply 1/3, you meant 33/100)

- 5 years ago

The max. value of ab + bc + ca is 0.

- 5 years, 1 month ago

that's wrong

- 5 years ago

maximum is never 0

- 5 years, 1 month ago

Not always true

- 5 years ago

E.g. Maximum of -x^2 where $$x \in \mathbb{R}$$ is obviously 0, attained when x = 0.

- 5 years ago