Maximum value.

hmm

can you pls tell me when will the equality hold ?

when x=1 y=0
x=-1 y=0

ok

how can u say this

=2(sinA-cosA) + sin2A(sinA-cosA)/2
it will attain max. value when sinA-cosA is max.

?

W.L.O.G

x≥y

so |x-y| + |x³- y³| = x-y + x³- y³

so maxima will occour when x>0 and y<0, let x= a>0, -y= b>0

so we have to maximize a+ a³+ b+ b³, when a²+ b² = 1 (*)

W.L.O.G let a≥b

so (1+ a²) ≥ (1+ b²)

By chebycheffs inequality

a(1+a²)+ b(1+ b²)2 ≤ (a+b)2(1+a²+ 1+ b²)2

and also , by AM-GM , (a+b)²4≤ a²+ b²2

Giving (a+b) ≤√2 , combine this with (*) , to get the answer as 3/√2, equality when x=y= 1/√2

Somehow I am getting a different answer. I may have goofed up but:

\because \ (x^3-y^3) = (x-y)(x^2+xy+y^2) and (x^2+xy+y^2)\ge 0

x-y and x³-y³ have the same sign.

So WLOG x≥y and the expression is therefore(x-y)+(x^3-y^3) = (x-y)[1+(x^2+xy+y^2)] = (x-y)(2+xy)

Also xy = \frac{1-(x-y)^2}{2}

So the expression is \frac{t(5-t^2)}{2} where t=x-y≥0

Now, this is increasing in \left[0, \sqrt{\frac{5}{3}} \right] and decreasing thereafter.

So the max occurs when t= \sqrt{\frac{5}{3}} < \sqrt 2

and equals

\frac{5 \sqrt{5}}{3 \sqrt{3}}

Thanks bhatt Sir. (Right answer)

also Thanks Subhodip, Arnab for valuable efford.