consider 2 triangles like $\bigtriangleup ABC \quad and \quad \bigtriangleup \acute{A}\acute{B}\acute{C}$, which $S_{\bigtriangleup \acute{A}\acute{B}\acute{C}} \leq S_{\bigtriangleup ABC}$.(S stands for surface)
Prove that $$\overline{\acute{A}\acute{B}} * \overline{\acute{A}\acute{C}}*\overline{\acute{B}\acute{C}} \leq \overline{AB}*\overline{AC}*\overline{BC}$$ ($\overline{AB}$ stands for length of AB side.)
any response would be appreciated.
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In this form, the statement cannot be true (if I have understood it correctly). It would imply that if the two triangles have the same area, then the product of the side lengths would coincide. This is clearly not true (to be explicit, consider the trangles with vertices $(0,0)$, $(1,0)$ and $(p,1)$ for varying $p$).
If I understand the question correctly, its easy to find a counter-example. We can use $abc=4AR$ here, where $A$ is the area, $R$ is the circumradius. Thus, we have to prove $4A_1R_1\leq 4A_2 R_2$, or at least $R_1\leq R_2$, because we know that $A_1\leq A_2$.
Its easy to see why the last statement can't be always true. Small triangles with one large angle, can have quite big radius, as the following the figure shows:
Whereas some another acute triangle with the same area, would have a much smaller radius, as you may see in the figure. Both triangles have area of $3.05$, however one has a radius of $64.5$, wheareas the other has of $1.6$. Thus, you can see why this might not be true. Using this, you may find a particular counter-example yourself.