Prove $(a^2+b^2+c^2)(a+b-c)(b+c-a)(a+c-b)\leq abc(ab+bc+ac)$ Let $a,b,c$ are $3$ edge of a triangle. Prove $(a^2+b^2+c^2)(a+b-c)(b+c-a)(a+c-b)\leq abc(ab+bc+ac)$.  My try: I suppose $c=\min\{a,b,c\}$ but I don't know what next.
 A: I give a correct answer here (in the last one there was a wrong factor $4$ instead of the correct $16$). 
Basically I use here the following two facts: 
(a) Stewart’s theorem.-It is applied to triangles with a cevian. In the figure below we apply this theorem twice, with the cevian $d=OG$ in $\triangle DOB$ and with the median (which is a cevian of course) $BD$ in $\triangle ABC$.
(b) The fact that $A-M\ge G-M$, applied to
$$ \frac{\frac ab+\frac ba+\frac bc+\frac cb+\frac ac+\frac ca}{6}\ge \sqrt[6]{1\cdot 1\cdot 1}=1$$ hence $$\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}\ge 6\qquad (1)$$

Taking in account the formulas $$S=\sqrt{p(p-a)(p-b)(p-c)}=\frac {abc}{4R}$$
giving the area $S$ of a triangle $\triangle {ABC}$, the proposed inequality becomes
the equivalent $$\frac{a^2+b^2+c^2}{R^2}\le (a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)\qquad (2)$$
We’ll prove $(2)$.
We have $$h^2=R^2\frac{b^2}{4}\qquad (3)$$
$$\text{(Stewart)}\space R^2m+h^2(2m)=(m+2m)(d^2+2m\cdot m)$$ i.e.
$$R^2+2h^2=3d^2+6m^2\qquad (4)$$ $(3)$ in $(4)$ gives
$$3R^2-\frac{b^2}{2}=3d^2+6m^2\qquad (5)$$
$$\text{(Stewart again)}:\space\space c^2(\frac b2)+a^2(\frac b2)=b(3m)^2+\frac{b^2}{4}$$ i.e.
$$c^2+a^2=18m^2+\frac{b^2}{2}\qquad (6)$$
Eliminating $m$ in $(5)$ and $(6)$ one gets
$$9R^2-(a^2+b^2+c^2)=9d^2\ge 0\Rightarrow \frac{a^2+b^2+c^2}{R^2}\le 9$$
$$\color{green}{\text{is it}\space 9\le (a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)?}$$
If YES, we have finish. In fact it is; because of $(1)$ one has 
$$(a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)=\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}+3\ge 6+3$$ 
This ends the proof.
