Timeline for Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 9, 2013 at 14:32 | vote | accept | Jeel Shah | ||
| Nov 9, 2013 at 14:32 | comment | added | Jeel Shah | Ahh, Okay. I thought I had to say it at the start but I get it now and Is see where the non negative part comes in handy. Thanks! | |
| Nov 9, 2013 at 12:55 | comment | added | Cameron Buie | @gekkostate: Not quite. Instead, say something like "Since $(a-b)^2\ge0$ for all real $a,b,$ then $a^2-2ab+b^2\ge0,$ so $a^2+2ab+b^2\ge4ab,$ so...." You need to remember what it is that you're trying to prove. Also, don't forget to use the fact that $a,b$ are non-negative. That will be important in one of your steps. (Do you see where?) | |
| Nov 9, 2013 at 5:29 | comment | added | Jeel Shah | Ohh okay. So once I have done that, in my final copy, I would write something like "It must be shown that $(a-b)^2 \ge 0$" or something like that and then I show my proof right? | |
| Nov 9, 2013 at 5:27 | comment | added | angryavian | @gekkostate By doing the work that you have done above! You can think of what you have done as scratch work, and then write your final solution starting with $(a-b)^2 \ge 0$. It may seem strange because you are starting with something out of the blue, but that's ok. | |
| Nov 9, 2013 at 5:25 | comment | added | Jeel Shah | This might a dumb question but how would you know that you have to start of with $|a-b| \ge 0$? | |
| Nov 9, 2013 at 5:21 | history | answered | Cameron Buie | CC BY-SA 3.0 |