1Hmmm ...... 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
Ricky
·2010-06-14 09:07:52
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 .
1
Ricky
·2010-06-14 09:08:11
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 .
341Wow! 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
1good observation prophet
i have done it a bit different way,.,.
341
Hari Shankar
·2010-06-15 00:12:47
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
62This one is a lovely proof.. awesome :)
souvik .. what is your proof?
1
souvik seal
·2010-06-15 02:51:46
RICKY u r trapped.......sum is made for that mistake.....
xy≤1
means -xy≥-1
8-6xy≥2
not ≤
