HELPPP MEE please
- Thread starter loca95
- Start date
Really? For a second I thought I misspoke when I tried to do it myself. Good on you if you found it.
Here is a shorter way if you know the arithmetic-geometric mean inequality:
\(\displaystyle \displaystyle \frac{a_1+a_2+\cdots +a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}\)
Then
\(\displaystyle \displaystyle \frac{2a+b}{3} = \frac{a+a+b}{3} \ge \sqrt[3]{a\cdot a\cdot b} = \sqrt[3]{a^2b}\)
Here is a shorter way if you know the arithmetic-geometric mean inequality:
\(\displaystyle \displaystyle \frac{a_1+a_2+\cdots +a_n}{n} \ge \sqrt[n]{a_1a_2\cdots a_n}\)
Then
\(\displaystyle \displaystyle \frac{2a+b}{3} = \frac{a+a+b}{3} \ge \sqrt[3]{a\cdot a\cdot b} = \sqrt[3]{a^2b}\)
yeah this is a very good way thank you so much and i m sorry for disturbance
can i ask another question?
Yes, but in the future please post your attempt at a solution and state where you may be stuck
Yes, that looks good, but the proof should be written backwards, unless you turn the implications into double-implications: \(\displaystyle \iff\)
ok thaaaaaanks a lot
D
Deleted member 4993
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This proof will be very difficult to write "backwards".
Instead follow the logic provided by daon with arithmatic mean and geometric mean.