PART-A
To show: ((a+b)/2)^2 >= ab
This is just another way of writing the well-known A.M. >= G.M.
Suppose you are not sure how to do it - one way is to restructure it until you can handle it, and then work backwards. You can liken this process to Reverse Engineering.
Above is equivalent to:
1) (a+b)^2 >= 4ab
2) a^2+2ab+b^2 >= 4ab
3) a^2-2ab+b^2 >= 0
4) (a-b)^2 >= 0 which we know is true for all real a,b
So you can reverse your steps and do the proof like this:
(a-b)^2 >= 0 ==> (a-b)^2 + 4ab >= 4ab ==> (a+b)^2 >= 4ab ==> [(a+b)/2]^2 >= ab
PART-B
^4 = \left( \frac{\frac {a+b}{2} + \frac {c+d}{2}}{2} \right )^2 \times \left (\frac {\frac {a+b}{2} + \frac {c+d}{2}}{2} \right )^2 \geq )
(\frac {c+d}{2}) \times (\frac {a+b}{2})(\frac {c+d}{2}) = \left (\frac {a+b}{2} \right )^2 \times \left (\frac {c+d}{2} \right )^2 \geq ab \times cd = abcd)
To show: ((a+b)/2)^2 >= ab
This is just another way of writing the well-known A.M. >= G.M.
Suppose you are not sure how to do it - one way is to restructure it until you can handle it, and then work backwards. You can liken this process to Reverse Engineering.
Above is equivalent to:
1) (a+b)^2 >= 4ab
2) a^2+2ab+b^2 >= 4ab
3) a^2-2ab+b^2 >= 0
4) (a-b)^2 >= 0 which we know is true for all real a,b
So you can reverse your steps and do the proof like this:
(a-b)^2 >= 0 ==> (a-b)^2 + 4ab >= 4ab ==> (a+b)^2 >= 4ab ==> [(a+b)/2]^2 >= ab
PART-B
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