# 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

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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.**