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anirudh alameluvari Grade: Upto college level
```        if a3 + b3 = a-b show that a2 + b2 < 1 if a,b are +ve reals
```
8 years ago

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.

```
8 years ago
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