Prove that $\sum\limits_{\mathrm{cyc}}\sqrt{xy^2(y+z)} \le {1 \over 2} \sqrt{(x + y + z)^4 - (x^2 + y^2 + z^2)^2}$ 
Prove the following inequality for all positive real numbers $x$, $y$, and $z$:
$$\sqrt{xy^2(y+z)} + \sqrt{yz^2(x+z)} + \sqrt{zx^2(x+y)} \le {1 \over 2} \sqrt{(x + y + z)^4 - (x^2 + y^2 + z^2)^2}$$

My attempt. I tried to prove it myself but I didn't know how to. I tried to transform this inequality but made it even more complicated. Is there an easy way to prove it? Help would be very appreciated.

Let's raise both sides of the inequality to the second power. Indeed, we just need to show:
$$2(\sqrt {xy^2(y+z)}\sqrt {yz^2(x+z)}+\sqrt {yz^2(x+z)}\sqrt {zx^2(x+y)}+\sqrt {xy^2(y+z)}\sqrt {zx^2(x+y)})\le x^3y+xz^3+y^3z+x^2y^2+x^2z^2+y^2z^2+2x^2yz+2xy^2z+2xyz^2.$$
Now, notice that: $$2\sqrt {xy^2(y+z)}\sqrt {yz^2(x+z)}\le yz[x(y+z)+y(x+z)]=2xy^2z+xyz^2+y^2z^2.$$
Hence, it suffices to prove that:
$$x^2yz+xy^2z+xyz^2\le x^3y+xz^3+y^3z; $$
and, this is almost obvious because;
$$(x^2yz+xy^2z+xyz^2)^2\le (x^3y+xz^3+y^3z)(xyz^2+xy^2z+x^2yz).$$
