1same as
xx-yx≥xy-yy
im done if i can prove that for all a>b and p>q,
ap-bp>aq-bq
Prove that for any " x , y > 0 " ,
x x + y y ≥ x y + y x
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9 Answers
21Consider f(x,y) = xx+ yy- xy-yx
f(x,y) = f(y,x) , the function is symmetric
Let us fix some 'y'(y>0) and consider p(x) = xx+ yy- xy-yx
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) = xx+ yy- xy-yx
we have proved f(x,y) ≥0 with equality iff x=y
1
1Could 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 ?
21lol sir i m unable to prove that p(x)≥0
but i m sure p(x)≥0 is it wrong?
341that would be the problem statement itself :D. you wouldnt need that f is symmetric in x,y
341that would be the problem statement itself :D. you wouldnt need that f is symmetric in x,y
21no i hav fixed 'y'
i want to use calculus to prove that p(x)≥0
and p(x) takes minima at x = y
262this is much simpler...
WAM >= WGM
xx + yy /2 >= √(xx .yy)
xy + yx >= √(xy. yx)
now,
(x.x + y.y) /2 >= (x+y)√(xx.yy)
(x.y +y.x )/2 >= √(xy.yx)
now we know, (x-y)2 >=0
or x2 +y2 >= 2xy
comparing with previous results we directly have the inequality