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Show that

$$\sqrt[n]{a_1a_2...a_n}\le\frac1n\sum_{k=1}^na_k$$

I have proved it for $n=2$ by defining

$$b=a_1-\frac{a_1+a_2}2$$

which means that $$-b=a_2-\frac{a_1+a_2}2$$

then rewriting the equation for $n=2$

$$a_1a_2 \le \left(\frac{a_1+a_2}2\right)^2$$

then substituting $b$ into the right hand side

$$a_1a_2=\left(\frac{a_1+a_2}2+b\right)\left(\frac{a_1+a_2}2-b\right)$$

$$a_1a_2= \left(\frac{a_1+a_2}2\right)^2-b^2$$

and as $b^2$ is always positive then $\displaystyle a_1a_2 \le \left(\frac{a_1+a_2}2\right)^2$must be true

However this is when I run into the problem of how to prove this for $n+1$, any help would be greatly appreciated

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Given $a_1,a_2,\ldots,a_{n+1}\geq 0$, let $A:=\frac{a_1+a_2+\ldots+a_{n+1}}{n+1}$. Define $b_i:=\sqrt{a_i\,A}$ for $i=1,2,\ldots,n-1$ and $b_n:=\sqrt{a_{n-1}a_n}$. Apply AM-GM for $n$ variables on the $b_i$'s, and then apply AM-GM on $2$ variables on each $b_i$.

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