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anirudh alameluvari Grade: Upto college level
        

if a+ b3 = a-b show that a2 + b2 < 1 if a,b are +ve reals


 

7 years ago

Answers : (1)

AskIITians Expert PRASAD IIT Kharagpur
18 Points
										

a,b positive reals so


a3 + b3 > 0    ,   so a - b > 0   so,  a > b.


As  a > 0, b > 0,     so, (a+b) > (a - b )   so ,  ((a-b) / (a + b)) < 1.


As,  a3 + b3 = a - b    so, a - a3  =  b + b3    so,  a/b = 1 + b2 / 1 - a2


Sy componendo-dividendo,    we have   (a+b)/(a-b) = (2 - a2 + b2) / (a2 + b2 )


so,  (a+b)/(a-b)  > 1,  we have   (2 - a2 + b2) > (a2 + b2 )which gives,     a2 < 1  so  0 <a <1,  as, b < a, so 0 < b < a < 1


so, ab<1.


a3 + b3  = a - b     given,


so, (a + b)(a2 + b2 - ab) = (a-b)


so,  a2 + b2 - ab  =   (a-b) / (a + b)


from above we have Right Hand Side less than 1,


so, a2 + b2 - ab < 1   so,  a2 + b2 < ab


 


As, ab < 1,


we have,           a2 + b2 < 1.


Hence Proved.


 

7 years ago
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