SOLUTION: if a,b are positive numbers such that a+b=1 , prove that a^2+b^2 >= 1/2

Algebra ->  Inequalities -> SOLUTION: if a,b are positive numbers such that a+b=1 , prove that a^2+b^2 >= 1/2      Log On


   

Question 1016895: if a,b are positive numbers such that a+b=1 , prove that a^2+b^2 >= 1/2
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
a%5E2+%2B+b%5E2++=+a%5E2+%2B+%281-a%29%5E2+=+2a%5E2+-2a%2B1
Now for any quadratic expression Ax%5E2+%2BBx+%2B+C, the maximum or minimum value is C-B%5E2%2F%284AC%29. In this case, A = 2 >0, and hence we have a minimum value.
The minimum value of 2a%5E2+-2a%2B1 is then equal to 1-%28-2%29%5E2%2F%284%2A2%2A1%29+=+1+-+4%2F8+=+1%2F2.
This ends the solution.