Prove that: #(a+b)/2 = sqrt( a*b)# When #a>=0# and #b>=0# ?

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Answer 1 · 2017-10-01T08:46:29

#(a+b)/2 color(red)(>=) sqrt(ab)" "# as shown below

Note that:

#(a-b)^2 >= 0" "# for any real values of #a, b#.

Multiplying out, this becomes:

#a^2-2ab+b^2 >= 0#

Add #4ab# to both sides to get:

#a^2+2ab+b^2 >= 4ab#

Factor the left hand side to get:

#(a+b)^2 >= 4ab#

Since #a, b >= 0# we can take the principal square root of both sides to find:

#a+b >= 2sqrt(ab)#

Divide both sides by #2# to get:

#(a+b)/2 >= sqrt(ab)#

Note that if #a != b# then #(a+b)/2 > sqrt(ab)#, since then we have #(a-b)^2 > 0#.