<s>[INST] 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.
 [/INST] 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;
$$(x^2yz+xy^2z+xyz^2)^2\le (x^3y+xz^3+y^3z)(xyz^2+xy^2z+x^2yz).$$
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