Question

Let a, b, c be three distinct positive real numbers in G.P., then $a^2+2bc-3ac$ is-

• A. > 0
• B. < 0
• C. = 0
• D. Can't be found out

Answer 1

The correct option is A > 0

Since $a,b,cϵR^{+}$ and distinct $\to AM>GM>HM$
Since $b=\sqrt{ac}$ and consider AM and HM of a and c,
$\Rightarrow \frac{a+c}{2}>b>\frac{2ac}{a+c}$
From first inequality $(a+c)>2b\Rightarrow a^2+ac-2ab>0$
From second inequality $b(a+c)>2ac$
$\Rightarrow 2ab+2bc-4ac>0$
Adding the two inequalities $a^2+2bc-3ac>0$.