Waste less time on Facebook — follow Brilliant.
×

Inequality mania!

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

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

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

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

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

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

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 \(x^3+y^3+z^3+6xyz\)

Mark Gilbert - 1 year, 9 months ago

Log in to reply

Wrote very nicely+1

Lakshya Sinha - 1 year, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...