all is in the title take two triangles of sides $a,b,c$ and $x,y,z$ so we have :
\begin{align} & (ab)^2x^2z^2+(ab)^2y^2z^2+(ac)^2x^2y^2+(ac)^2y^2z^2+(bc)^2y^2x^2+(bc)^2(x^2z^2) \\[10pt] \leq {} & a^4(yz)^2+z^4(ab)^2+(ac)^2y^4+b^4(xz)^2+(bc)^2x^4+c^4(xy)^2 \end{align}
I have a method but it's very ugly so I would like to know a better method.
My question : What is your fastest method ?
Edit : This is my proof after some computation we find :
\begin{align} & a^4(yz)^2+z^4(ab)^2+(ac)^2y^4+b^4(xz)^2+(bc)^2x^4+c^4(xy)^2-[(ab)^2x^2z^2+(ab)^2y^2z^2+(ac)^2x^2y^2+(ac)^2y^2z^2+(bc)^2y^2x^2+(bc)^2(x^2z^2)] = {} [a^2(y^2+z^2-x^2)+b^2(-y^2+z^2+x^2)+c^2(x^2+y^2-z^2)]^2-(a+b+c)[\prod_{cyc}(a+b-c)](x+y+z)\prod_{cyc}(x+y-z) & \end{align}
Wich is just Neuberg Pedoe inequality . But it's very ugly .
Thanks .
We need to prove that $$\frac{a^2b^2}{xy}\cdot\frac{x^2+y^2-z^2}{xy}+\frac{a^2c^2}{xz}\cdot\frac{x^2+z^2-y^2}{xz}+\frac{b^2c^2}{yz}\cdot\frac{y^2+z^2-x^2}{yz}\leq\frac{a^4}{x^2}+\frac{b^4}{y^2}+\frac{c^4}{z^2}.$$ Let $\xi$, $\psi$ and $\eta$ be measures-angles of the triangle $XYZ$ with sides-lengths $x$, $y$ and $z$
and let $\frac{a^2}{x}=p$, $\frac{b^2}{y}=q$ and $\frac{c^2}{z}=r$.
Thus, we need to prove that
$$2pq\cos\eta+2pr\cos\psi+2qr\cos\xi\leq p^2+q^2+r^2.$$ Now, let $\vec{p}\upuparrows\vec{XY}$ and $|\vec{p}|=p$,
$\vec{q}\upuparrows\vec{YZ}$ and $|\vec{q}|=q$
and $\vec{r}\upuparrows\vec{ZX}$ and $|\vec{r}|=r$.
Id est, we need to prove that $$\left(\vec{p}+\vec{q}+\vec{r}\right)^2\geq0,$$ which is obvious.
Done!