If $a,b,c$ are in $A.P.$ then prove that $b^2>ac$

Question

If $a,b,c$ are in $A.P.$ then prove that $b^2>ac$

My Attempt:

$a,b,c$ are in AP $$b=\dfrac {a+c}{2}$$

Also, $$A.M>G.M$$ $$\dfrac {a+b}{2}>\sqrt {ab}$$

Answer 1

Since $b = (a+c)/2$, we have

$\begin{array}\\ b^2-ac &=(a+c)^2/4-ac\\ &=(a^2+2ac+c^2)/4-ac\\ &=(a^2-2ac+c^2)/4\\ &=(a-c)^2/4\\ &\ge 0\\ \end{array} $

with equality if and only if $a = c$.

Answer 2

Straightforward: b=a+x, c=a+2x, $ac=a(a+2x)=a^2+2ax\lt b^2=(a+x)^2=a^2+2ax+x^2$.

Answer 3

Well, presumabaly $b = a + d=c-d;$ and $c = a + 2d$.

So $b^2 = (c-d)(a+d) = ac +(c-a)d - d^2 = ac + 2d*d -d^2 = ac + d^2$.