# Showing an inequality with ln I have to show that the following inequation is true: (ln(x) + ln(y))/(2) <= ln((x+y)/(2))

Showing an inequality with ln
I have to show that the following inequation is true:
$\frac{\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)}{2}\le \mathrm{ln}\left(\frac{x+y}{2}\right)$
I transformed it into
$\frac{\mathrm{ln}\left(x\cdot y\right)}{2}\le \mathrm{ln}\left(x+y\right)-\mathrm{ln}\left(2\right)$
because I thought that I better can show the inequation here, but I don't know how to proceed.
How can I proceed or am I completely wrong?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Tirioliwo
How about we exponentiate both sides? We get
${e}^{\left(ln\left(x\right)+ln\left(y\right)\right)/2}={e}^{ln\left(x\right)/2}{e}^{ln\left(y\right)/2}=\sqrt{xy}$
and
${e}^{ln\left(\left(x+y\right)/2\right)}=\left(x+y\right)/2$
Now the result is immediate per the AM-GM inequality.