Waste less time on Facebook — follow Brilliant.
×

Most of CMI's questions are proof questions. Here's one more:

If p, q, r are real numbers such that \(p^2+q^2+r^2=1\), then prove that

\(3p^2q+3p^2r+2q^3+2r^3 \leq 2\).

Note by Vishnu C
2 years, 7 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

We simplify the given expression as follows -

\[\begin{eqnarray} 3p^2q+3p^2r+2q^3+2r^3 &=& 3p^2(q+r)+2(q+r)(q^2-qr+r^2) \\ &=& (q+r)(3p^2+2q^2+2r^2-2qr) \\ &=& (q+r)(3(p^2+q^2+r^2)-(q^2+r^2)-2qr) \\ &=& (q+r)(3-(q+r)^2) \\ &=& 3(q+r)-(q+r)^3 \\ \end{eqnarray}\]

Now since, \(q,r \in \mathbb{R} \), then let \(s=q+r\) and thus \(s \in \mathbb{R}\). We must obviously have \[3p^2q+3p^2r+2q^3+2r^3 = 3s-s^3 \leq 2\]

Kishlaya Jaiswal - 2 years, 7 months ago

Log in to reply

What inequality is \(3s-s^3 \leq 2\)?

Vishnu C - 2 years, 7 months ago

Log in to reply

Consider \(f(s) = 3s-s^3\).

We have \(f^\prime(s) = 3-3s^2 = 0 \Rightarrow s=\pm1\). Observe that \(f^{\prime\prime}(1) <0\) and so we should have maximum at \(s=1\). And hence \(f(s)_{\text{max}}=2\)

And therefore, \(3s-s^3 \leq 2\)

Kishlaya Jaiswal - 2 years, 7 months ago

Log in to reply

@Kishlaya Jaiswal Doh!

Vishnu C - 2 years, 7 months ago

Log in to reply

@Vishnu C Thanks!

Kishlaya Jaiswal - 2 years, 7 months ago

Log in to reply

@Kishlaya Jaiswal I'm going to look like a complete idiot for saying this, but this is the truth: I got all the way up to \((q+r)(3-(q+r)^2)\) in my simplification. And then I simply left the question. I didn't know how to proceed any further. DAMN IT!!!

Vishnu C - 2 years, 7 months ago

Log in to reply

Very nice proof @Kishlaya Jaiswal . It boils down to recognizing that the expression is a function of \( q + r \), from which the result follows.

Calvin Lin Staff - 2 years, 7 months ago

Log in to reply

Thank You Sir.

Actually at first sight, I was thinking if we can manipulate the expression so as to form function in \(p^2,q^2,r^2\) so that we can use the condition \(p^2+q^2+r^2=1\) alongwith AM-GM Inequality. But that didn't worked. It was then I realized that it's a function in \(q+r\)

Kishlaya Jaiswal - 2 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...