sides of two triangles which have different areas 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.

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: !enter image description here

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.