An ancient statement, and I once read an elementary proof but I don't recall. Who can help me? The origin of the inequality is a regular magazine for high school teachers issued by P. Noordhoff since 1914 (Nieuw Tijdschrift voor Wiskunde).
I considered various methods to prove the statement above. For instance using the fact that integers $n$ which are of the form $x^2+xy+y^2$ for two relatively prime integers $x,y$, are precisely those positive integers occurring as divisors of $m^2+m+1$ for some integer $m$.
Another approach I tried is to consider the general solutions $x,y$ of $n=x^2+xy+y^2.$ An integer $n$ can be uniquely written as $n=sz^2$ with $s$ squarefree. Let
$$\begin{align*} s=f^2+fg+g^2,&\quad z=m^2+mn+n^2;\\ x=(f+g)n^2+2gmn-fm^2,&\quad y=-gn^2+2fnm+(f+g)m^2\,. \end{align*}$$
Then we have $x^2+xy+y^2=(f^2 + fg + g^2)(m^2 + mn + n^2)^2$. The radical of $xy(x+y)(x^2+xy+y^2)$ has the form $xy(x+y)sz$, which leads to the (interesting) identity
$$\bigl(xy(x+y)\bigr)sz\bigl(fg(f+g)\bigr)=sz\prod_{i=1}^3{(u_i^2-sn^2)}\,,$$
where $u_1=gn-fm$, $u_2=(f+g)m+fn$, and $u_3=gm+(f+g)n$.
I also tried to use the following summation identity: let $e>0$ be an integer and $w=(-1)^e$. Choose $r=\frac{1}{3}(2^e-w)$ which is an integer. For $f,g,m,n$ define
$$x=\frac14\sum_{i=0}^r\sum_{j=0}^2\binom{2^e}{3i+j}n^jm^{-j}(-1)^in^{3i}m^{2^e-3i}\,\times\\ \bigr[(f+(f + 2g)w)j^2+(-f + 2g +(-5f - 4g)w)j -2f - 2g +(2f - 2g)w\bigl]\,,$$
$$y=\frac14\sum_{i=0}^r\sum_{j=0}^2\binom{2^e}{3i+j}n^jm^{-j}(-1)^in^{3i}m^{2^e-3i}\,\times\\ \bigr[(-f-g-(-f+g)w)j^2+(3f+g -(-f - 5g)w)j + 2g -(4f+2g)w\bigl]\,.$$
Then the equality $x^2+xy+y^2=(f^2+fg+g^2)(m^2+mn+n^2)^{2^e}$ holds, which enable us to find all solutions with $s_i$ square-free of the form $s_0\prod_k{s_k^{2k}}$.
N.B.: Please note that the transformation $x\mapsto x+y,y\mapsto-y$ yields the same value for $x^2+xy+y^2$.
Why of interest? A corollary of $(x+y)^2<\operatorname{rad}\bigl[xy(x+y)(x^2+xy+y^2)\bigr]$ is that for $ABC$ triples with $A+B=C$ and $\operatorname{rad}(ABC)<C$, we have $A^2+AB+B^2=C^2-AB$. Thus,
$$\operatorname{rad}(C^2-AB)<\leq C^2-AB\operatorname{rad}(ABC)\operatorname{rad}(C^2-AB)-AB<C\operatorname{rad}(C^2-AB)-AB\,,$$
which implies $AB<(C-1)\operatorname{rad}(C^2-AB)\,.$ We can also conclude that $C=A+B<\operatorname{rad}(C^2-AB)$, and thus for small values of $\operatorname{rad}(C^2-AB)=3,7,13,19,21\dots$ (OEIS sequence A034017) there are only a limited number of ABC triples.