How to prove that if $x>0$ and $y>0$, then $\sqrt{x}+\sqrt{y}>\sqrt{x+y}$, using the relation of arithmetic and geometric means? 
How to prove that if $x>0$ and $y>0$, then $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}\,,$$ using the relation of arithmetic and geometric means?

I started by showing that if $x>0$ and $y>0$, based on the relation of arithmetic and geometric means, $\dfrac{x+y}{2}\ge\sqrt{xy}$.
Hence, $x+y\ge2\sqrt{xy}$.
I am now stuck here and don't know what must be the next step.
Any suggestions or comments will be much appreciated.


Theorem (A Baby Version of the Triangle Inequality).    Let $a$, $b$, $c$, and $d$ be real numbers.  We have
$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}\,.$$
The equality holds if and only if

*

*$(a,b)=(0,0)$, or

*there exists $\lambda\geq 0$ such that $(c,d)=(\lambda a,\lambda b)$.


Let $x$ and $y$ be nonnegative real numbers.  Note that
$$\sqrt{x}+\sqrt{y}=\sqrt{\sqrt{x}^2+0^2}+\sqrt{0^2+\sqrt{y}^2}\,.$$
By the Baby Triangle Inequality above,
$$\sqrt{\sqrt{x}^2+0^2}+\sqrt{0^2+\sqrt{y}^2}\geq \sqrt{(\sqrt{x}+0)^2+(0+\sqrt{y})^2}\,.$$
That is,
$$\sqrt{x}+\sqrt{y}\geq \sqrt{x+y}\,.\tag{#}$$
By the equality conditions of the Baby Triangle Inequality, (#) is an equality if and only if $x=0$ or $y=0$.

Now, I shall prove the Baby Triangle Inequality using the AM-GM Inequality to fulfill the OP's request that the AM-GM Inequality must be used.  By squaring the required inequality, what we need to prove is equivalent to
$$\sqrt{a^2+b^2}\sqrt{c^2+d^2}\geq ac+bd\,.$$
We, in fact, have a stronger inequality:
$$\sqrt{a^2+b^2}\sqrt{c^2+d^2}\geq |a|\,|c|+|b|\,|d|\,. \tag{*}$$
By squaring the inequality above, we know that (*) is equivalent to
$$a^2d^2+b^2c^2\geq 2\,|a|\,|b|\,|c|\,|d|\,,$$ which is true by the AM-GM Inequality:
$$\frac{a^2d^2+b^2c^2}{2}=\frac{|ad|^2+|bc|^2}{2}\geq \sqrt{|ad|^2\cdot |bc|^2}=|a|\,|b|\,|c|\,|d|\,.$$

