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](http://books.google.nl/books/about/Nieuw_tijdschrift_voor_wiskunde.html?id=t5TxAAAAMAAJ)).

I considered various methods to prove the statement above. For instance using the [fact](https://mathoverflow.net/questions/78361/which-integers-take-the-form-x2-xy-y2) 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](http://www.emis.de/journals/GM/vol13nr2/andrica/andrica.pdf) $x,y$ 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.