if anyone do not post the solution between 3 days i will post it>>>>classic inequality

Hmmm ...... Lagrange Multipliers will surely kill it -

We first formally set ,

g ( x , y ) = k ( x + y - 2 )

f ( x , y ) = x ³ y ³ ( x ³ + y ³ )

And , h ( x , y ) = f ( x , y ) + g ( x , y ) .

Here , " k " is actually a constant not equal to zero .

Now , we shall find out the partial derivatives of " h ( x , y ) " each with respect to " x " , " y " ,

and " k " and then equate them to zero simulataneously .

∂ h∂ x = 6 x ⁵ y ³ + 3 x ² y ⁶ + k = 0 ...... ( 1 )

∂ h∂ y = 6 y ⁵ x ³ + 3 y ² x ⁶ + k = 0 ...... ( 2 )

∂ h∂ k = x + y - 2 = 0 ...... ( 3 )

1 >

x + y = 0 ------- Obviously this violates the conditions already set , that " x + y = 2 " .

2 >

x - y = 0 ------ This implies x = y = 1 .

So f ( 1 , 1 ) = 2

3 >

x y = 0 -------This obviously holds only when x = 0 , y = 2 ; or x = 2 , y = 0 .

These values clearly give ,

f ( 0 , 2 ) = f ( 2 , 0 ) = 0 .

So studying all the cases , we see that the function " f ( x , y ) " attains its maximum value at x = y = 1 .

That value is , 2 .

Hence , we proved the inequality .

Wow! That's some heavy machinery coming into play.

But we can use AM-GM to finish this easily.

x^3y^3(x^3+y^3) = (x+y)x^3y^3 [(x+y)^2-3xy] = 2x^3y^3 [4-3xy]

Hence we need to prove that x^3y^3 (4-3xy) \le 1

From AM-GM 1 = \frac{xy + xy + xy + 4-3xy}{4} \ge ( x^3y^3 (4-3xy))^{\frac{1}{4}}

\Rightarrow x^3y^3 (4-3xy) \le 1

good observation prophet
i have done it a bit different way,.,.

I remembered I had encountered this about 2 years ago at http://www.goiit.com/posts/list/algebra-answer-to-reddevil-s-inequality-qn-76299.htm

That solution is a bit more involved.

Edit: We need to observe from AM-GM that xy≤1/4 which i what allows us to take 4-3xy as +ve