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if a3 + b3 = a-b show that a2 + b2 < 1 if a,b are +ve reals
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.
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