×

# $$x, y, z$$

Let $$x, y, z$$ be three positive integers such that $$x+y+z=1$$. Show that $x^5\sqrt{x}+y^5\sqrt{y}+z^5\sqrt{z}\leq\frac{1}{10}\sum\limits_{i=2}^{10}\left(x^i+y^i+z^i+\frac{1}{9}\right )$

Note by Jaydee Lucero
3 years, 11 months 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:

x,y ,z positive reals?

It follows from AM-GM: $$\sum_{i=1}^{10} x^i \ge 10 x^5\sqrt{x}$$. Then just add three together.

- 3 years, 11 months ago

Comment deleted Dec 12, 2013

$$\sum_{i=2}^{10}(x^i+y^i+z^i + \frac{1}{9}) = \sum^{10}_{i=2}(x^i+y^i+z^i ) + 1 = \sum^{10}_{i=1}(x^i+y^i+z^i)$$

- 3 years, 11 months ago

sorry, brainfart, failed to realize that the $$x + y + z$$ is also multiplied by $$\frac{1}{10}$$, silly me.

- 3 years, 11 months ago

yeah.. precisely!

- 3 years, 11 months ago

if the base of the isosceles triangle joints the points (2,-4), (1,-3); the area is 9/2 . find the third vertex of the triangle.