Proof that $ 2\sqrt{x_1x_2} \le x_1+x_2 $

Question

We have $$ x_1, x_2,...,x_n \ge 0 $$

and $$ P(n): x_1x_2....x_n \le \left(\frac{x_1+...+x_n}{n}\right)^{n} $$

I have to prove that P(2) is valid.

$$x_1 x_2 \le \left(\frac{x_1+x_2}{2}\right)^{2} $$

I don't know how to achieve this, this is what I tried so far:

$$ \sqrt{x_1x_2} \le \frac{x_1+x_2}{2} $$

$$ 2\sqrt{x_1x_2} \le x_1+x_2 $$

Here, I don't know how to go further. Is this a good way proving this or am I completely wrong?

Answer 1

Hint: $(a - b)^2 \ge 0$. Expand and manipulate

Answer 2

It's $$\left(\sqrt{x_1}-\sqrt{x_2}\right)^2\geq0.$$

Answer 3

The important parts about the proof are already given in the answers up there. I want to share something different, geometrical actually (not that it works for a proof though) Consider a circle $OAB$ (the center is $T$) with a diameter $|AB|$ now inside the circle choose a point $P$ ($P\in|AB|$)such that $|AP|=x_1$ and $|BP|=x_2$ and then choose a point $K$ on the circle such that $<KPB=90^\circ$

As we know from Euclid's theorem $|KP|=\sqrt{x_1x_2}$ and as we know that if this segment $|KP|$ tends to the center $T$ it will be closer and closer to the radius which is $\dfrac{x_1+x_2}{2}$ so we can conclude that it is smaller or equal to the radius, Thus;

$$\sqrt{x_1x_2}\leq\frac{x_1+x_2}{2}$$

As I've said this is completely partial, although fun If you draw the shape and see. This also works only for two variables.