Monday, September 15, 2014

let a,b,c,d are positive real number such that a.b=c.d and a-b>c-d. Prove that a+b>c+d

We have 4 positive real numbers a, b, c and d such that
a*b = c*d and a - b > c - d.


If (a - b) > (c
- d)


=> (a - b)^2 > (c -
d)^2


=> a^2 - 2ab + b^2 > c^2 - 2cd +
d^2


it is given that ab =
cd


=> a^2 - 2ab + b^2 + 4ab > c^2 - 2cd + d^2
+ 4cd


=> a^2 + 2ab + b^2 > c^2 + 2cd +
d^2


=> (a + b)^2 > (c +
d)^2


=> (a + b) > (c + d) as a, b, c and d
are positive and so are a + b and c + d.


This
proves that given a*b = c*d and a - b > c - d, we have a + b >c +
d.

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