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 )

The values of “ x “ and “ y “ so found from these 3 equations are those values of “ x “ and “ y “ for

which “ f ( x ) “ is maximized .

If multiple values are possible , then we have to find out the functional values in each case and

compare them manually .

Subtracting ( 2 ) from ( 1 ) ,

6 x ⁵ y ³ - 6 y ⁵ x ³ = 3 y ² x ⁶ - 3 x ² y ⁶

or , 2 x ³ y ³ ( x ² - y ² ) = x ² y ² ( x ⁴ - y ⁴ ) = x ² y ² ( x ² - y ² ) ( x ² + y ² )

or , x ² y ² ( x ² - y ² ) ( x ² + y ² - 2 x y ) = 0

or , x ² y ² ( x - y ) ² ( x + y ) = 0

Now comes three cases .

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

RICKY u r trapped.......sum is made for that mistake.....
xy≤1
means -xy≥-1
8-6xy≥2
not ≤