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!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWe 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.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

@Julian Yu

Log in to reply