×

# Absolute 2

Let $$a, b$$ and $$c$$ be positive real numbers such that $$a + b + c \leq 4$$ and $$ab + bc + ca \geq 4$$.

Prove that at least two of the inequalities

$$|a - b| \leq 2, |b - c| \leq 2, |c - a| \leq 2$$

are true.

Note by Sharky Kesa
3 years, 9 months 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:

For sake of contradiction let us suppose $$|a-b| > 2$$, $$|b-c| > 2$$,$$|c-a| > 2$$. which implies $(a-b)^{2} + (b-c)^{2} + (c-a)^{2} > 8$ ---(1)

Given $$a + b + c \le 4$$ $a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca \le 16$ $a^{2} + b^{2} + c^{2} - ab - bc - ca \le 16 - 3ab -3bc - 3ca$ $a^{2} + b^{2} + c^{2} - ab - bc - ca \le 4$ $2a^{2} + 2b^{2} + 2c^{2} - 2ab - 2bc - 2ca \le 8$ $(a-b)^{2} + (b-c)^{2} + (c-a)^{2} \le 8$.

This contradicts (1) and hence our desired proof follows.

- 3 years, 9 months ago

its not true

- 3 years, 9 months ago

It actually is true.

- 3 years, 9 months ago