AM-GM Problems

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Eugenis

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Eugenis

#1 h Apr 28, 2015, 1:04 AM • 2 Y

Y by Adventure10, Mango247

Can somebody give me some AM-GM problems?

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WorstIreliaNA

451 posts

WorstIreliaNA

#2 h Apr 28, 2015, 1:05 AM • 1 Y

Y by Adventure10

If $x$ is a positive real number, find the minimum of $x+\frac{1}{x^2}$

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chezbgone

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chezbgone

#3 h Apr 28, 2015, 1:08 AM • 2 Y

Y by Adventure10, Mango247

Here are some (simpler) problems:
http://www.artofproblemsolving.com/community/q1h568395p3333294

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Eugenis

2404 posts

Eugenis

#4 h Apr 28, 2015, 1:12 AM • 2 Y

Y by Adventure10, Mango247

The AM-GM inequality states that for a set of nonnegative real numbers $a_1,a_2,\ldots,a_n$ , the following always holds: $\frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n}$
For the set of real numbers given, it can be established that

$\frac{x^3+1}{2x^2} \ge \frac{\sqrt{x}}{x} \implies x^3+1 \ge 2x\sqrt{x}$
How to continue?

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Eugenis

2404 posts

Eugenis

#5 h Apr 28, 2015, 1:16 AM • 2 Y

Y by Adventure10, Mango247

Through several manipulations, I got that

$x^4+\frac{1}{x^2}-2x \ge 0$

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AMCprep

1407 posts

AMCprep

#6 h Apr 28, 2015, 1:19 AM • 3 Y

Y by max-, Adventure10, Mango247

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

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BOGTRO

5818 posts

BOGTRO

#7 h Apr 28, 2015, 1:19 AM • 5 Y

Y by Eugenis, Imayormaynotknowcalculus, Ultroid999OCPN, Adventure10, Mango247

Usually AM-GM works best when the RHS is a constant, which motivates writing $x+\frac{1}{x^2}=\frac{1}{2}x+\frac{1}{2}x+\frac{1}{x^2}$ . Can you see how to continue from here?

Also quite important is that equality holds only when $a_1=a_2=\hdots=a_n$ , so you couldn't have split it up as $\frac{1}{3}x+\frac{2}{3}x+\frac{1}{x^2}$ or something.

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MathSlayer4444

1631 posts

MathSlayer4444

#8 h Apr 28, 2015, 1:23 AM • 2 Y

Y by Adventure10, Mango247

Hint 1
If that hint wasn't enough,
Hint 2

This post has been edited 1 time. Last edited by MathSlayer4444, Apr 28, 2015, 1:23 AM

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Eugenis

2404 posts

Eugenis

#10 h Apr 28, 2015, 1:36 AM • 2 Y

Y by Adventure10, Mango247

BOGTRO, I don't understand why you would assume that they are equal though...

This post has been edited 1 time. Last edited by Eugenis, Apr 28, 2015, 1:41 AM

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tkhalid

965 posts

tkhalid

#11 h Apr 28, 2015, 1:39 AM • 1 Y

Y by Adventure10

Eugenis wrote:

The AM-GM inequality states that for a set of nonnegative real numbers $a_1,a_2,\ldots,a_n$ , the following always holds: $\frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n}$
For the set of real numbers given, it can be established that

$\frac{x^3+1}{2x^2} \ge \frac{\sqrt{x}}{x} \implies x^3+1 \ge 2x\sqrt{x}$
How to continue?

as you wrote in your post Eugenis

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Eugenis

2404 posts

Eugenis

#12 h Apr 28, 2015, 1:40 AM • 1 Y

Y by Adventure10

Ah, referring to my post...

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Wave-Particle

3690 posts

Wave-Particle

#13 h Apr 28, 2015, 1:43 AM • 1 Y

Y by Adventure10

AMCprep wrote:

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

Shouldn't it be the maximum? Because isn't the product on the right hand side?

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YanYau

133 posts

YanYau

#14 h Apr 28, 2015, 1:44 AM • 4 Y

Y by Eugenis, rjiang16, Adventure10, Mango247

The AMGM inequality states that:

$\frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n}$

With equality if and only if $a_1=a_2=\hdots=a_n$

So when $a_1=a_2=\hdots=a_n$ :

$\frac{a_1+a_2+\ldots+a_n}{n}=\sqrt[n]{a_1a_2\cdots a_n}$

When you are asked to find the maximum/minimum value of an expression, you also have to find the equality case, so you need to make sure you split the terms in a way that the equality case can be achieved.

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tkhalid

965 posts

tkhalid

#15 h Apr 28, 2015, 1:46 AM • 2 Y

Y by Adventure10, Mango247

For your problem, if you've managed to get it down to showing $x^4+\frac{1}{x^2}-2x\geq 0$ , then just add $2x$ and use AM-GM

Also for problems see the attachment, but a heads up: The solutions are directly below the problems, so you might wanna cover them up.

Attachments:

Chapter 25 AM-GM Inequalities.pdf (111kb)

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tkhalid

965 posts

tkhalid

#16 h Apr 28, 2015, 1:47 AM • 2 Y

Y by Adventure10, Mango247

anandiyer12 wrote:

AMCprep wrote:

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

Shouldn't it be the maximum? Because isn't the product on the right hand side?

Thats the cool thing about the question

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Eugenis

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Eugenis

#17 h Apr 28, 2015, 1:47 AM • 2 Y

Y by Adventure10, Mango247

tkhalid, I got the same result but don't know how to continue.

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Eugenis

2404 posts

Eugenis

#18 h Apr 28, 2015, 1:48 AM • 2 Y

Y by Adventure10, Mango247

By the way, I will work the problems in your handout. Thanks!

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YanYau

133 posts

YanYau

#19 h Apr 28, 2015, 1:49 AM • 1 Y

Y by Adventure10

tkhalid wrote:

anandiyer12 wrote:

AMCprep wrote:

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

Shouldn't it be the maximum? Because isn't the product on the right hand side?

Thats the cool thing about the question

Here's a hint

This post has been edited 1 time. Last edited by YanYau, Apr 28, 2015, 1:53 AM

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tkhalid

965 posts

tkhalid

#20 h Apr 28, 2015, 1:52 AM • 1 Y

Y by Adventure10

Well I didn't actually get that result, I just assumed you did that part right. So if it is right, then we have $x^4+\frac{1}{x^2}\geq 2\sqrt{x^4\cdot \frac{1}{x^2}}=2\sqrt{x^2}=2x$ . So subtracting $2x$ gives $x^4+\frac{1}{x^2}-2x\geq 0$ .

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tkhalid

965 posts

tkhalid

#21 h Apr 28, 2015, 1:53 AM • 1 Y

Y by Adventure10

YanYau wrote:

Here's a hint

I think you meant $1+a\geq 2\sqrt{a}$ .

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YanYau

133 posts

YanYau

#22 h Apr 28, 2015, 1:53 AM • 1 Y

Y by Adventure10

whoops yup, edited

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lminsl

544 posts

lminsl

#23 h Apr 28, 2015, 9:10 AM • 2 Y

Y by Adventure10, Mango247

Find the maximum value of the function
$x(1-x^n)$

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bluehall90

134 posts

bluehall90

#24 h Apr 28, 2015, 10:15 AM • 2 Y

Y by YanYau, Adventure10

Inequalities : A Mathematical Olympiad Approach by Radmila Bulajich Manfrino, José Antonio Gómez Ortega, and Rogelio Valdez Delgado in AM-GM chapter is a very good introduction in inequalities imo. The exercise are ranged from the easy one to a difficult one. I love working with it.

This post has been edited 1 time. Last edited by bluehall90, Apr 28, 2015, 10:32 AM
Reason: grammar fix

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cobbler

2180 posts

cobbler

#26 h Apr 28, 2015, 11:05 AM • 2 Y

Y by Adventure10, Mango247

Whoops, my last hint was a typo, it was meant to say Click to reveal hidden text

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rctmathadventures

685 posts

rctmathadventures

#27 h Sep 2, 2018, 6:30 AM • 2 Y

Y by Adventure10, Mango247

is the minimum $3(1/4)^{1/3}$

This post has been edited 1 time. Last edited by rctmathadventures, Sep 2, 2018, 6:31 AM

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Knty2006

50 posts

Knty2006

#28 h Nov 17, 2019, 10:46 PM • 1 Y

Y by Adventure10

AMCprep wrote:

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

By Holders Inequality, $(1+a)(1+b)(1+c)(1+d) \geq (1+abcd)^4=16$

This post has been edited 1 time. Last edited by Knty2006, Nov 15, 2020, 12:03 PM

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akasht

83 posts

akasht

#30 h May 23, 2020, 2:05 AM

Y by

Knty2006 wrote:

AMCprep wrote:

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

By Holders Inequality, $(1+a)(1+b)(1+c)(1+d) \geq (1+abcd)^4=8$

Shouldn’t it be 16?

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MathJams

3229 posts

MathJams

#31 h May 23, 2020, 2:59 AM

Y by

I suggest going on Alcumus. There are many good problems there!

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alphaone001

963 posts

alphaone001

#32 h May 23, 2020, 3:29 AM

Y by

AMCprep wrote:

Here's a really nice problem:

$a,b,c,d$ are positive reals and $abcd=1$
$(1+a)(1+b)(1+c)(1+d)$ - find the minimum value.

Solution

@below oops haha

This post has been edited 1 time. Last edited by alphaone001, May 23, 2020, 4:52 AM
Reason: i am such a smart kid

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Bowser498

1743 posts

Bowser498

#33 h May 23, 2020, 4:18 AM

Y by

alphaone001 wrote:

Solution

FTFY since the inequalities involve $b+1 \geq 2\sqrt{b}$ , $c+1 \geq 2\sqrt{c}$ , and $d+1 \geq 2\sqrt{d}$ , and their product must be $16\sqrt{abcd}$ .

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Vndom

351 posts

Vndom

#34 h May 23, 2020, 5:36 PM

Y by

Eugenis wrote:

Can somebody give me some AM-GM problems?

You can try Alcumus.

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