If \(a,b,c\) are positive reals and \(n\) is a positive integer , Prove the following inequality:

\[\large\dfrac{na}{b+nc} + \dfrac{nb}{c+na} +\dfrac{nc}{a+nb} \geq \dfrac{3n}{n+1}\]

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:

TopNewestProve \[{\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}}\geq{\frac{3abc}{ab+bc+ca}} \]

Log in to reply

\(a,b,c\) are positive reals right?

Log in to reply

Yes. It is a nice problem :) But there is a stronger one after.

Log in to reply

Log in to reply

Log in to reply

Click here :)

Log in to reply

\[\large {\sum_{cyc} \dfrac{a^3}{a^2+ab+b^2} \\ = \sum_{cyc} \dfrac{a^4}{a(a^2+ab+b^2)} \\ \geq \dfrac{\left(\displaystyle\sum_{cyc} a^2\right)}{\displaystyle\sum_{cyc} a(a^2+ab+b^2)} \\ = \dfrac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)} \\ = \dfrac{a^2+b^2+c^2}{a+b+c} \\ \geq \dfrac{(a+b+c)^2}{3(a+b+c)} \\ = \dfrac{a+b+c}{3} }\]

Thus it suffices to prove that:

\(\large{\dfrac{a+b+c}{3} \geq \dfrac{3abc}{ab+bc+ac} \Rightarrow (a+b+c)(ab+bc+ac) \geq 9abc}\)

Proof:

\[\large{(a+b+c)(ab+bc+ac) \\ = \sum_{cyc} (a^2b+abc+a^2c) \\ = 3abc + \sum_{cyc} a^2(b+c) \\ = 3abc+abc\left(\sum_{cyc} \dfrac{a(b+c)}{bc}\right) \\ = 3abc+abc\left(\sum_{cyc} \left(\dfrac{a}{b}+\dfrac{b}{a}\right)\right) \\ \geq 3abc+6abc = 9abc} \]

Log in to reply

Wrong for \( n = -1\) :P

Log in to reply

Thanks edited.

Log in to reply

By T2's Lemma \[\sum_{cyc}\dfrac{na}{b+nc}\ge \dfrac{n^2(a+b+c)^2}{(ab+bc+ca)(n^2+n)}=\dfrac{n(a+b+c)^2}{(n+1)(ab+bc+ca)}\ge \dfrac{3n}{n+1}\] done.

Log in to reply

Here's a strengthening (albeit not a very good one):

Given that \(a^2+b^2+c^2=1\), prove that \[\dfrac{na}{b+nc}+\dfrac{nb}{c+na}+\dfrac{nc}{a+nb}\ge \dfrac{3n}{n+3(a^3b+b^3c+c^3a)}\]

Log in to reply

Done the same way. And by the way,the inequality was not first found by Titu (neither by Arthur Engel),it was found by some other Russian mathematician. It is mentioned in "Kvant".

Log in to reply

Titu is such a cute name lol

Log in to reply

Log in to reply

That is true, but still most people call it T2's lemma or Engel form of CS.

Log in to reply

Yay! I got inspired by this note quite lucid it is.

Log in to reply

@Calvin Lin @Harsh Shrivastava @Alan Yan @Saarthak Marathe Hope all enjoy solving it :)

Log in to reply