Question

prove that if \( 0 < a < b \) then \[a < \sqrt{ab} < \frac{a+b}{2} < b\]

Answer 1

\[(a+b)/2\] is average value of a and b so obviously it will be between a and b

\[\sqrt{ab}\] since a is smaller than b

\[\sqrt{ab}<\sqrt{b*b}=b\]

\[\sqrt{ab}>\sqrt{a*a}=a\]

Answer 2

it remains to show that \( \sqrt{ab} \leq \frac{a+b}{2},\ \ \ a,b \geq 0\)

Answer 3

thinking.....

Answer 4

square them both
\[\frac{a^2}{4}+\frac{a b}{2}+\frac{b^2}{4}>\text{ab}\]

Answer 5

equivalent to \[a^2+b^2>2ab\]

Answer 6

\[a^2+b^2-2ab>0\]
\[(a-b)^2>0\]
which is true.

Answer 7

If \(a < b \) then \[a = \frac{a+a}{2} < \frac{a+b}{2} < \frac{b+b}{2} < b\] now if \( 0 < a < b\) then \( a^2 < ab\) so \( a < \sqrt{ab}\)\[(a-b)^2 > 0 \implies \]\[a^2 + b^2 >2ab\]\[a^2 + 2ab + b^2 > 4ab\]\[(a+b)^2 > 4ab\]so\( a+b > 2 \sqrt{ab} \). Moreover, for all a,b, we have \( (a-b)^2 \geq 0 \), and thus \( (a+b)^2 \geq 4ab\)
which implies that \( a+b \geq 2\sqrt{ab} \) for all \(a,b \geq 0 \)

Answer 8

looks good