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30 Jul 2008 13:50:48 IST

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Answer to reddevil's Inequality Qn

None

I am unable to post the answer in the same thread where reddevil has posted the question.

The inequality to be proved as I gather is

if x+y = 2, $x^3 y^3 (x^3+y^3) \le 2$ for positive real numbers x and y

By AM-GM, we have $1 = \frac{x+y}{2} \ge \sqrt{xy} \Rightarrow xy \le 1$

Let xy = a.

Then, $0 \le a \le 1$

We have,

x^3y^3(x^3+y^3) = (xy)^3 [(x+y)^3 - 3xy(x+y)] = a^3(8-3a) = 8a^3 - 6a^4

f(a) = 8a^3 - 6a^4

Hence, we can easily see that the maximum is attained when a=1 and equals 2.

Hence $x^3y^3(x^3+y^3) \le 2$ when x+y = 2

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Comments (6)

Conjurer

Blazing goIITian

Joined: 20 Feb 2008

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30 Jul 2008 15:03:11 IST

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Quoting from the other post

As I mentioned before x^3+y^3>= 2xy ( Look at image)

So x^3 y^3 (x^3 + y^3) >= 2x^4 y^4

And by AM GM max of x^4y^4 = 1, which gets us x^3 y^3 (x^3 + y^3) >= 2 (I

am not sure about this)

PS: Consider this also x^2 + y^2 >= 2xy

So x^2 + y^2 - xy >= xy

Mutiplying both sides by (x+y)

So x^3 + y^3 >= 2xybut take 3/2 and 1/2 this doesnt hold, I have simply used simple algebra. What wrong?

(x^3 + y^3) > = (x+y)xy

Hari Shankar

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30 Jul 2008 15:39:01 IST

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The result you got by appying Cauchy Schwarz is same as that you get by using Rearrangement Inequality

$x^3+y^3 \ge x^2y+y^2 x = xy(x+y) = 2xy (\because x+y=2)$

Hence $x^3y^3(x^3+y^3) \ge 2x^4y^4$

The next step however is logically incorrect. a>b and b<c, does not allow us to conclude that a<c.

As to part 2, I dont see why it does not hold for x=3/2 and y = 1/2

Conjurer

Blazing goIITian

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30 Jul 2008 18:45:58 IST

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Oops I guess I was miscalculating 3/2 and 1/2 all this while.

OK I get your point but how does it contradict that x^3 y^3 (x^3 + y^3) >=2 , I mean when the product xy is maximum then also this inequality is true isnt it?

Hari Shankar

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30 Jul 2008 18:52:21 IST

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yeah, it is actually equal at that point. not greater than (check that)

Conjurer

Blazing goIITian

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30 Jul 2008 18:54:45 IST

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Thank you sir.

Madmax

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4 Aug 2008 21:12:04 IST

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very well solved sir

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