The question starts off by saying, show that $\ln(1+x) \leq x$ which is easy enough.
But then it says, by making an appropriate choice of x, hence prove that:
$(a_1a_2\cdot\cdot\cdot a_n)^{1/n} \leq \frac{a_1+a_2+\cdot\cdot\cdot+a_n}{n}$
The way I tried it is letting $x=a_1a_2…a_n-1$
Then,
$\ln(a_1a_2\cdot\cdot\cdot a_n) \leq a_1a_2\cdot\cdot\cdot a_n-1$
Multiplying both sides by $\frac{1}{n}$
$\ln((a_1a_2…a_n)^{\frac{1}{n}}) \leq \frac{a_1a_2\cdot\cdot\cdot a_n-1}{n}$
Raising both sides to e
$(a_1a_2…a_n)^{\frac{1}{n}} \leq e^{\frac{a_1a_2…a_n-1}{n}}$
We get the RHS of the wanted result but other than that everything is different.
Best Answer
Asserting $\ln(1+x)\leqslant x$ is the same thing as asserting that $x+1\leqslant e^x$, which, in term, is the same thing as saying that $x=(x-1)+1\leqslant e^{x-1}$.
Now, let $m=\frac{a_1+\cdots+a_n}n$. We may assume that each $a_k$ is greater than $0$. Then the previous inequality tells us that\begin{align}\frac{a_1}m\times\frac{a_2}m\times\cdots\times\frac{a_n}m&\leqslant\exp\left(\frac{a_1}m-1\right)\exp\left(\frac{a_2}m-1\right)\cdots\exp\left(\frac{a_n}m-1\right)\\&=\exp\left(\frac{a_1+\cdots+a_n}m-n\right)\\&=\exp(n-n)\\&=1.\end{align}This means, of course, that$$a_1a_2\cdots a_n\leqslant m^n,$$which, in turn, means that$$\sqrt[n]{a_1a_2\cdots a_n}\leqslant\frac{a_1+\cdots+a_n}n.$$