Waste less time on Facebook — follow Brilliant.
×

Need help on this (full solution)

\[\large P=\frac{1}{2+x+yz}+\frac{1}{2+y+xz}+\frac{1}{2+z+xy}\] If \(x,y\) and \(z\) are positive reals satisfying \(4(x+y+z)=3xyz\), find the maximum value of \(P \).

Note by Gurīdo Cuong
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

I think I've got it! The answer is 3/8.

\( \text{Let's switch variables to a, b and c, where, } \) \[a = \frac{1}{yz}, b = \frac{1}{xz}, c = \frac{1}{xy} \\ \] \( \text{And P becomes, after substituting,} \\ P = \dfrac{a}{2a + \frac{a^2}{\sqrt{abc}} + 1} + \dfrac{b}{2b + \frac{b^2}{\sqrt{abc}} + 1} + \dfrac{c}{2c + \frac{c^2}{\sqrt{abc}} + 1} \\ \text{The given condition becomes, } a + b + c = \frac{3}{4} \\~\\ \text{So much nicer, right? Now, AM-GM gives} \\ (abc)^{1/3} \leq \frac{(a + b + c)}{3} = \frac{1}{4} \\ \Rightarrow abc \leq \frac{1}{64} \Rightarrow \sqrt{abc} \leq \frac{1}{8} \Rightarrow 8a^2 \leq \frac{a^2}{\sqrt{abc}} \\~\\ \text{As this term and similar ones are in the denominator,} \\ P \leq \dfrac{a}{8a^2 + 2a + 1} + \dfrac{b}{8b^2 + 2b + 1} + \dfrac{c}{8c^2 + 2c + 1} \\~\\ \\~\\ \text{Now, let's look at one of these terms, } \\ \dfrac{a}{8a^2 + 2a + 1} = \dfrac{2a}{16a^2 + 4a + 2} = \dfrac{2a}{(4a - 1)^2 + 12a + 1} \leq \dfrac{2a}{12a + 1} = \dfrac{1}{6} \cdot \left(1 - \dfrac{1}{12a + 1}\right) \\~\\ \\~\\ \text{But, by Cauchy-Schwarz (Titu's Lemma), } \\ \dfrac{1^2}{12a + 1} + \dfrac{1^2}{12b + 1} + \dfrac{1^2}{12c + 1} \geq \dfrac{(1 + 1 + 1)^2}{12(a + b + c) + 3} = \dfrac{9}{9 + 3} = \dfrac{3}{4} \\~\\ \text{Thus, } \\ P \leq \dfrac{1}{6} \cdot \left(3 - \left(\dfrac{1}{12a + 1} + \dfrac{1}{12b + 1} + \dfrac{1}{12c + 1}\right) \right) \leq \dfrac{1}{6} \cdot (3 - \dfrac{3}{4}) = \dfrac{3}{8} \Rightarrow P \leq \dfrac{3}{8} \\ \text{And, equality holds everywhere, if } a = b = c = \frac{1}{4} \Rightarrow x = y = z = 2 \)

Ameya Daigavane - 1 year, 8 months ago

Log in to reply

I feel that it's correct.

Swapnil Das - 1 year, 8 months ago

Log in to reply

What is the level of this question? What is the minimum grade we should have studied in to understand this question?

Akhash Raja Raam - 1 year, 9 months ago

Log in to reply

Grade 10 mathematical olympiad

Ms Ht - 1 year, 8 months ago

Log in to reply

Oh...Thanks!!

Akhash Raja Raam - 1 year, 8 months ago

Log in to reply

@Akhash Raja Raam Also In Viet Nam student grade 9 who study specialized mathematics can solved this...

Ms Ht - 1 year, 8 months ago

Log in to reply

@Ms Ht They can but I am pretty much sure I have learned just enough in my 10th grade. I specialize in my field later. Thanks for the information!

Akhash Raja Raam - 1 year, 8 months ago

Log in to reply

3/8

Sumon Jose - 1 year, 9 months ago

Log in to reply

\(x=y=z=2\)

Ms Ht - 1 year, 9 months ago

Log in to reply

How do you know that?

Pi Han Goh - 1 year, 9 months ago

Log in to reply

Here is my solution. It has more of basic maths.

By applying AM-GM Inequality

xyz≥6

x+y+z≥8

Now the important thing to note here is that in P the variables are in the denominator. Also it is understood that if the denominator is big,the faction is small and vice versa. For the maximum value of P the denominators should be the smallest.

Now if we apply AM-GM inequality to 2+ x+yz We get,

2+x+yz ≥(6)³√2 Six times cube root of 2

This will be the same for the other terms

So P =1/(6³√2 ) + 1/(6³√2 ) + 1/(6³√2 )

= 1/(2³√2) 1 divided by two times cube root of two

≈0.396

And 0.396 > 0.375 which is 3/8

There you go bro...:) :)

Surya Sharma - 1 year, 9 months ago

Log in to reply

I've thought of this solution before, but you see the equality holds when \(x=y=z=xy=yz=xz=2\), which can't happened

Gurīdo Cuong - 1 year, 9 months ago

Log in to reply

Comment deleted Feb 23, 2016

Log in to reply

@Brilliant Member I'm talking about his solution. The way he used AM-GM led to \(x=y=z=xy=yz=xz=2\) when the equality holds

Gurīdo Cuong - 1 year, 9 months ago

Log in to reply

As pointed out by @Gurīdo Cuong, you "proved" incorrectly that \( P \geq \approx 0.396 \).

In particular, since \( P ( 2, 2, ,2 ) = 3.75 < 0.396 \), your solution is wrong.

Note: I do not know how you proved that \( x + y + z \geq 8 \), which is false like in the case of \( x = y = z = 2 \).

Calvin Lin Staff - 1 year, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...