Another Inequality

Consider f(x,y) = x^x+ y^y- x^y-y^x

f(x,y) = f(y,x) , the function is symmetric

Let us fix some 'y'(y>0) and consider p(x) = x^x+ y^y- x^y-y^x

for x>0 , it's easy to conclude using elementary calculus that p(x)≥0

p takes it minimum value i.e. zero at x = 'y'

similar argument holds for g(y) = x^x+ y^y- x^y-y^x

we have proved f(x,y) ≥0 with equality iff x=y

same as
x^x-y^x≥x^y-y^y

im done if i can prove that for all a>b and p>q,
a^p-b^p>a^q-b^q

Could you please show how you got " p ( x ) ≥ 0 " , Shubhodip ? Otherwise , it's alrighty :) .

@Gordo , exactly what I wanted . Could you please complete the solution ?

@gordo: you may hit rough weather when 0<a,b<1

lol sir i m unable to prove that p(x)≥0

but i m sure p(x)≥0 is it wrong?

that would be the problem statement itself :D. you wouldnt need that f is symmetric in x,y

that would be the problem statement itself :D. you wouldnt need that f is symmetric in x,y

no i hav fixed 'y'

i want to use calculus to prove that p(x)≥0

and p(x) takes minima at x = y

this is much simpler...

WAM >= WGM

x^x + y^y /2 >= √(x^x .y^y)

x^y + y^x >= √(x^y. y^x)

now,
(x.x + y.y) /2 >= _(x+y)√(x^x.y^y)

(x.y +y.x )/2 >= √(x^y.y^x)

now we know, (x-y)² >=0
or x² +y² >= 2xy

comparing with previous results we directly have the inequality