New user? Sign up

Existing user? Sign in

Let \(x,y,z \ge 0\) where \(x+y+z=1\). Prove that

\[ \large{x^3+y^3+z^3+6xyz \ge \dfrac{1}{4} }.\]

Note by Lakshya Sinha 1 year, 9 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

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

> 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"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

\(\frac{1}{4}=\frac{(x+y+z)^3}{4}\)

\(x^3+y^3+z^3+6xyz\geq\frac{(x+y+z)^3}{4}\)

\(4x^3+4y^3+4z^3+24xyz\geq x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+6xyz\)

\(3x^3+3y^3+3z^3+18xyz\geq3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2\)

\(x^3+y^3+z^3+6xyz\geq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2\)

Which is true by Schur's. Equality when two of \(x, y, z\) equal \(\frac{1}{2}\) and the third equals \(0\). Note that the original inequality would still hold if it were \(x^3+y^3+z^3+3.75xyz\) instead of \(x^3+y^3+z^3+6xyz\)

Log in to reply

Wrote very nicely+1

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:

TopNewest\(\frac{1}{4}=\frac{(x+y+z)^3}{4}\)

\(x^3+y^3+z^3+6xyz\geq\frac{(x+y+z)^3}{4}\)

\(4x^3+4y^3+4z^3+24xyz\geq x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+6xyz\)

\(3x^3+3y^3+3z^3+18xyz\geq3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2\)

\(x^3+y^3+z^3+6xyz\geq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2\)

Which is true by Schur's. Equality when two of \(x, y, z\) equal \(\frac{1}{2}\) and the third equals \(0\). Note that the original inequality would still hold if it were \(x^3+y^3+z^3+3.75xyz\) instead of \(x^3+y^3+z^3+6xyz\)

Log in to reply

Wrote very nicely+1

Log in to reply