Waste less time on Facebook — follow Brilliant.
×

True or false?

Are the following statements true?

a) For \(a,b,c> 0\), \[( a^{3}c + b^{3}a + c^{3}b)^{2} ≥ 3a^{2}b^{2}c^{2}(ab + bc + ca)\]

b) For \(k \in \mathbb N*\), \[\sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}} \in \mathbb Q\]

Note by Gabi Dobre
1 year, 8 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

Phew, another inequality, another wild ride!
 

Expanding the LHS,
\( (\sum\limits_{cyc}^{} a^3c)^2 = \sum\limits_{cyc}^{} a^6c^2 + 2\sum\limits_{cyc}^{} a^4b^3c\)

By AM - GM, \( a^6c^2 + b^6a^2 \geq 2a^4b^3c \)
Adding similar inequalities, \( \sum\limits_{cyc}^{} a^6c^2 \geq \sum\limits_{cyc}^{} a^4b^3c \)

\(\text{LHS} \geq 3\sum\limits_{cyc}^{} a^4b^3c \geq 3\sum\limits_{cyc}^{} a^3b^3c^2 = \text{RHS}\), by Muirhead.
If Muirhead is too uncomfortable/artificial for you, I'll show you another way to prove \(\sum\limits_{cyc}^{} a^4b^3c \geq \sum\limits_{cyc}^{} a^3b^3c^2 \)
 

The inequality is equivalent to,
\( \sum\limits_{cyc}^{} \dfrac{a^2b}{c} \geq \sum\limits_{cyc}^{} ab \)
Now, by AM-GM, \( \dfrac{4}{7}\dfrac{a^2b}{c} + \dfrac{1}{7}\dfrac{c^2a}{b} + \dfrac{2}{7}\dfrac{b^2c}{a} \geq ab \)
Adding similar inequalities for the other two terms, we are done.

Ameya Daigavane - 1 year, 8 months ago

Log in to reply

b)
\(\displaystyle\;\;\;\;\sqrt{1+\frac 1{k^2}+\frac 1{(k+1)^2}}\)
\(\displaystyle=\sqrt{\frac{k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}}\)
\(\displaystyle=\sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}\)
\(\displaystyle=\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}\)
\(\displaystyle=\frac{k^2+k+1}{k(k+1)}\)
\(\in \mathbb Q\)

展豪 張 - 1 year, 8 months ago

Log in to reply

I think you should try a). Its more difficult.

Gabi Dobre - 1 year, 8 months ago

Log in to reply

I know...so I try b) first, haha! I'm still thinking about it......

展豪 張 - 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...