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

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?

• AM/GM${{{{}}}}$ Oct 25, 2017 at 18:54
• You are assuming what you have to prove. How about looking at the difference of the two sides of your proposed inequality? Oct 25, 2017 at 18:56
• $(x_1 - x_2)^2 \geq 0$. The LHS is $x_1^2 - 2x_1 x_2 + x_2^2$. Add $4x_1 x_2$ to both sides to obtain $x_1^2 + 2x_1 x_2 + x_2^2 \geq 4x_1 x_2$. Simplify the LHS and divide by $4$.
– user169852
Oct 25, 2017 at 18:57
• @Bungo I don't see why $$x_1^{2}+2x_1x_2+x_2^{2} \ge 4x_1x_2$$ is valid when dividing both sides by 4 ? Oct 25, 2017 at 19:06
• @Blnpwr You can always divide both sides of an inequality by a positive number. For example, if $a \geq b$ then $a-b$ is nonnegative, hence so is $(a-b)/4 = a/4 - b/4$, and therefore $a/4 \geq b/4$.
– user169852
Oct 25, 2017 at 19:10

It's $$\left(\sqrt{x_1}-\sqrt{x_2}\right)^2\geq0.$$
Hint: $(a - b)^2 \ge 0$. Expand and manipulate
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;
$$\sqrt{x_1x_2}\leq\frac{x_1+x_2}{2}$$