harder 3 Unit Inequalities question (2 Viewers)

tywebb · Jan 3, 2012

Omnipotence said:
Didn't Andrew Wiles prove the Modularity Theorem (Taniyama-Shimura) rather than directly Fermat's?

Andrew Wiles only proved the semistable case of the Taniyama-Shimura conjecture. It was not necessary to prove the full Taniyama-Shimura conjecture in order to prove Fermat's Last Theorem. Here is his proof: http://users.tpg.com.au/nanahcub/flt.pdf

Fermat's Last Theorem results as follows:

The full Taniyama-Shimura conjecture (now called the modularity theorem) was proved by Christophe Breuil, Brian Conrad, Fred Diamond, and Richard Taylor in 1999. Here is their proof: http://math.stanford.edu/~lekheng/flt/bcdt.pdf

asianese · Jan 3, 2012

Actually with the AM-GM inequality you should not just assume it but prove it before hand. The questions like this should be made so that you are made to prove it before anyway but still...

They're all quite simple - provided you don't take the long arguous route for the 3VAR case.

Also questions like the one above are quite standard inequalities. You should memorise such simpler proofs - obviously with knowing what's going on.

tywebb · Jan 3, 2012

There are many ways to prove the AM-GM inequality, but my favourite is Pólya's Dream. Pólya actually dreamt it!

$\textstyle A\ge G\text{ where } A={\sum_{i=1}^nx_i\over n}\text{ and } G=\root n\of{\prod_{i=1}^nx_i}$

$\textstyle{\bf Proof.}\text{ If } f(x)=e^x-x-1\text{ then }f'(x)=e^x-1=0\text{ where }x=0\text{ and }f''(x)=e^x>0\text{ for all real }x.\text{ So }f(x)\text{ has a global minimum of 0 when }x=0\text{ and therefore }e^x\ge x+1.\text{ Letting }x={x_i\over A}-1,\newline{\prod_{i=1}^nx_i\over A^n}=\prod_{i=1}^n{x_i\over A}=\prod_{i=1}^n(1+{x_i\over A}-1)\le\prod_{i=1}^ne^{{x_i\over A}-1}=e^{{\sum_{i=1}^nx_i\over A}-n}=e^{{nA\over A}-n}=1$

$\textstyle\therefore A^n\ge\prod_{i=1}^nx_i\therefore A\ge\root n\of{\prod_{i=1}^nx_i}\therefore A\ge G.\ \ \ \ \ \square$

ihave2shadows · Oct 30, 2014

Another method to prove
1/a + 1/b + 1/c >= 9 is to show a+1/a>=2 and similarly a/b+1/(a/b)>=2
(1/a + 1/b + 1/c)(a+b+c)=a/a+a/b+a/c+a/b+b/b+b/c+c/a+b/c+c/c>=3+2+2+2>=9

harder 3 Unit Inequalities question (2 Viewers)

tywebb

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