×

# 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
2 years, 1 month 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:

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

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

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

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

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

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 $$x^3+y^3+z^3+6xyz$$

- 2 years, 1 month ago

Wrote very nicely+1

- 2 years, 1 month ago