×

# Inequalities Help!

Given $$a, b, c$$ are positive real numbers such that $$ab+bc+ca=\frac { 1 }{ 3 }$$, prove the following inequality:

$\large \frac { a }{ { a }^{ 2 }-bc+1 } +\frac { b }{ { b }^{ 2 }-ca+1 } +\frac { c }{ { c }^{ 2 }-ab+1 } \ge \frac { 1 }{ a+b+c }$

Any help would be appreciated, thanks!

Note by Julian Yu
1 year, 10 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:

We have ,

$$a^{2} -bc + 1 = a(a + b + c) + \frac{2}{3} \small \text{ [Since ab+bc+ca = 1/3]}$$

Similarly $${ b }^{ 2 }-ca+1 = b(a+b+c) + \frac{2}{3}$$ and $$c^{2} -ab+1 = c(a+b+c) + \frac{2}{3}$$

Therefore our inequality becomes $$\large \frac { 1}{ (a + b + c) + \frac{2a}{3} } +\frac { 1 }{ (a + b + c) + \frac{2b}{3}} +\frac { 1 }{ (a + b + c) + \frac{2c}{3}} \ge \frac { 1 }{ a+b+c }$$

Now applying Titu's Lemma , $\large \frac { 1}{ (a + b + c) + \frac{2a}{3} } +\frac { 1 }{ (a + b + c) + \frac{2b}{3}} +\frac { 1 }{ (a + b + c) + \frac{2c}{3}} \ge \dfrac{(1+1+1)^{2}}{5(a+b+c)}$

Now it is obvious that $$\dfrac{(1+1+1)^{2}}{5(a+b+c)} \ge \dfrac { 1 }{ a+b+c }$$

since a,b,c are positive reals.

- 1 year, 10 months ago

the denominator at the end of the titu's lemma calculation is wrong it should be 11/3 (a+b+c) @Julian Yu @Harsh Shrivastava

- 1 year, 9 months ago

@Harsh Shrivastava Thanks, but the numerators are a,b,c not 1,1,1.

- 1 year, 10 months ago

I divided both numerator and denominator by a,b,c.

- 1 year, 10 months ago

Oh okay thanks!

- 1 year, 10 months ago