Is this a legal way to prove an inequality? I have to prove the following inequality:
$(x+y)\sqrt{\frac{x+y}{2}}\geq x\sqrt{y}+y\sqrt{x}$ where $x,y>0$.
After squaring both sides I obtain:
$(x^2+2xy+y^2)\frac{(x+y)}{2}\geq x^2y+xy^2$
then I simplify to
$x^2+y^2\geq 0$.
But this is always true. So my question is does this prove the inequality and if yes why equality is never achieved?!
 A: Summary answer: No, it is not a proof.
It may be a good way to find a proof, but after you do this, you need to check that the steps work backward.  
Sometimes they don't work backward.  Example:
PROVE $-2 > -1$: Square both sides, get $4>1$, which is true, done. ???
A: Equality is achieved for $x = y$. 
Basically this is a valid method for developing the idea for a proof, but not necessarily for actually proving the inequality: you've 'followed your nose' from the inequality you want to prove to an inequality that's easier to understand. What you then need to check is that all those steps work in reverse as well (i.e. that you can start with an inequality that you know to be true, and can work forward to show that implies the desired inequality). 
Equivalently, if you work backward through the proof like this, you can just make sure that all the implications are "if and only if", and then there's no need to 'redo' things.^*
In this case, you've made mistakes in your calculations: for instance $(x\sqrt y + y \sqrt x)^2 \neq x^2y + xy^2$ as you state. Further, 
$$
(x^2 + 2xy + y^2)\dfrac{1}{2}(x + y) = \frac12 x^3 + \frac32xy^2 + \frac32 x^2 y + \frac12y^3
$$
*: an example, for real numbers $a,b$ 


*

*$a \leq b \Leftrightarrow e^a \leq e^b$ 

*$a \leq b \leq c \Rightarrow a\leq c $  but  $a \leq c \not\Rightarrow a \leq b \leq c$
Basically, when you're using inequalities to imply other inequalities, take some care in thinking about whether you are dealing with equivalences or not.
A: Well here's I think a better proof. Use AM-GM inequality for the square root part and you have:
$$(x+y)\sqrt{\frac{x+y}2} \ge (x+y)\sqrt[4]{xy}$$
Now if we prove that:
$$(x+y)\sqrt[4]{xy} = x^{\frac54}y^{\frac 14} + y^{\frac 54}x^{\frac 14}\ge x\sqrt{y} + y\sqrt{x}= xy^{\frac 12} + yx^{\frac 12}$$
we're done. This follows from Muirhead's inequality, and here's the proof. Divide both sides $x^{\frac 14}y^{\frac 14}$ and we have:
$$x + y \ge x^{\frac34}y^{\frac14} + y^{\frac34}y^{\frac14}$$
Which follows from adding the AM-GM inequalities:
$$\frac{x+x+x+y}{4} \ge \sqrt[4]{x^3y} \quad \text {and} \quad \frac{x+y+y+y}{4} \ge \sqrt[4]{xy^3}$$
Additionally you can prove that the inqeuality holds when $x=y$ using the AM-GM properties
