+1 vote
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Let $x_1 > x_2>0$. Then which of the following is true?

1. $\log \big(\frac{x_1+x_2}{2}\big) > \frac{\log x_1+ \log x_2}{2}$
2. $\log \big(\frac{x_1+x_2}{2}\big) < \frac{\log x_1+ \log x_2}{2}$
3. There exist $x_1$ and $x_2$ such that $x_1 > x_2 >0$ and $\log \big(\frac{x_1+x_2}{2}\big) = \frac{\log x_1+ \log x_2}{2}$
4. None of these

recategorized | 32 views

+1 vote

Here $x_1>x_2>0 \Rightarrow x_1 \ne x_2 \Rightarrow (x_1-x_2)^2 \ne 0$.

and we know from simple algebra that

$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2\tag{i}$

Now $\forall~ x_1, x_2 \in \mathbb{R}$ and $x_1 \ne x_2$, we obtain

\begin{align} (x_1-x_2)^2 &>0 \\ \Rightarrow (x_1+x_2)^2-4x_1x_2 &>0 ~;~[\mathrm{From~no(i)}] \\ \Rightarrow (x_1+x_2)^2 &> 4x_1x_2 \\ \Rightarrow x_1+x_2 &>2\cdot\sqrt{x_1x_2}\\ \Rightarrow \frac{x_1+x_2}{2}&> (x_1x_2)^{\frac{1}{2}} \\ \Rightarrow \log\left( \frac{x_1+x_2}{2} \right) &> \log(x_1x_2)^{\frac{1}{2}} \\ \Rightarrow \log\left( \frac{x_1+x_2}{2} \right) &> \frac{1}{2}\log(x_1x_2)~;~[\because \log a^r=r\log a] \\ \Rightarrow \log\left( \frac{x_1+x_2}{2} \right) &> \frac{\log(x_1)+\log(x_2)}{2} ~;~[\because \log (xy)=\log x + \log y] \end{align}

So the correct answer is A.

by Active (3.2k points)