# Show that radical(1+x) < 1+ ( 1/2)*x if x>0 show steps

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Start by squaring both sides. The square of radical (1+x) is going to be (1+x). The square of 1+(1/2)*x will be 1+x+x^2.

Then subtract 1+x from both sides, which should be unproblematic as x>0. You will be left with 0 on the left hand side and x^2 on the right, and if x>0, then 0<x^2.

The radical ,(1+x) could be written in the index form like: (1+x)^(1/2).

Let a = (1+x/2) and b = (1+x)^(1/2).

Now to prove that a-b is positve or > 0.

Consider a^2-b^2 = (a-b)(a+b) = (1+x/2)^2-{(1+x)^(1/2)]^(2) = {(1+x/2)-[(1+x)^(1/2)}{(1+x/2)+(1+x)^1/2) .But this is postive as this is also equal to (1+x/2)^2-(1+x)^(1/2)^2=1+x+x^2/4-(1+x) =x^2/4, a positive quantity.

Therefore,{(1+x/2)-[(1+x)^(1/2)}{(1+x/2)+(1+x)^(1/2)=x^2/4 >0 or dividing bothe sides, by (1+x/2)+(1+x)^1/2), which is obviously +ve , when x>0, we get:

(1+x) - (1+x/2)^1/2) > 0. So it is proved.