![In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if ,a. BC = EF ,b. AC = DE ,c. AC = In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if ,a. BC = EF ,b. AC = DE ,c. AC =](https://d138zd1ktt9iqe.cloudfront.net/media/seo_landing_files/in-triangles-abc-and-def-ab-fd-and-a-d-1645598005.png)
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if ,a. BC = EF ,b. AC = DE ,c. AC =
![inequality - Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a( ac)^b(ab)^c}$ - Mathematics Stack Exchange inequality - Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a( ac)^b(ab)^c}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/7VVVL.jpg)
inequality - Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a( ac)^b(ab)^c}$ - Mathematics Stack Exchange
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