How to prove that if $x$ and $y$ are real numbers greater than $0$, then $\sqrt{x}+ \sqrt{y}>\sqrt{x+y}$?
Here is my attempt:
To show that $A>B$, we need to show that $A-B>0$. This implies that we need to show that $$\sqrt{x}+\sqrt{y}-\sqrt{x+y}>0.$$
I'm now stuck in this portion.
From $\sqrt{x}+\sqrt{y}-\sqrt{x+y}$, and using $\sqrt{x+y}\gt0$, we can write:
$$\sqrt{x}\sqrt{x+y}+\sqrt{y}\sqrt{x+y}-\sqrt{x+y}\sqrt{x+y}$$
$$=\sqrt{x(x+y)}+\sqrt{y(x+y})-(x+y)$$
As $\sqrt{x(x+y)}\gt\sqrt{x^2}=x$, and $\sqrt{y(x+y)}\gt\sqrt{y^2}=y$, we are done.
Alternatively, we know $x+y\lt x+y+2\sqrt{xy}=(\sqrt{x}+\sqrt{y})^2$, and take positive square roots.
let A=√x+√y and B=√(x+y) squaring A and B , we get $$ A^2=x+y+2*√x*√y$$ and $$B^2=x+y$$ since x,y>0, Therefore $$2*√x*√y>0$$ adding x+y both sides we get $$x+y+2*√x*√y>x+y$$ $$i.e., A^2>B^2$$ $$i.e., A>B$$ Since A,B>0
There is a trick in order to solve this type of inequalities.
You can try to manipulate the expression that you want to prove and get an obvious result that can be derived from your hypothesis.
$$ \sqrt{x}+\sqrt{y} > \sqrt{x+y} \Rightarrow (\sqrt{x}+\sqrt{y})^2 > (\sqrt{x+y})^2 $$
In this step I use the fact that for every pair of nonnegative real numbers $(a,b)$ it is true that if $a>b$ then $a^2>b^2$. So
$$ x +2\sqrt{xy} +y > x+y $$
Then
$$ 2\sqrt{xy} >0 $$
With this last inequality (that is certainly true if you have two positive numbers $x,y$), you can derive what you want to prove only by doing all that stuff that you did in order to get it but in a reverse way
Start with
$$ 2\sqrt{xy} >0 $$
then
$$ 2 \sqrt{xy} +x+y>x+y $$
so
$$ (\sqrt{x}+\sqrt{y})^2 > ((x+y)^{1/2})^2 $$
and you can "take" off your squares since $\sqrt{x}+\sqrt{y}$ and $(x+y)^{1/2}$ are positive real numbers.
