According to AM–GM inequality,
$\frac {a+b}{2}$ $\geq\sqrt{ab}$
$\Rightarrow$ $\frac{(a+b)^{2}}{4}$ $\geq ab$
$\Rightarrow (a+b)^{2}\geq 4ab$
$\because (a+b)^{2}\mid 4ab$
$\Rightarrow (a+b)^{2}\leq 4ab$
$\Rightarrow (a+b)^{2}=4ab$
$\therefore (a-b)^{2}=0$
$\therefore a=b$.
Search found 5 matches
- Fri Apr 29, 2016 12:29 pm
- Forum: Secondary Level
- Topic: Number Theory
- Replies: 5
- Views: 4780
- Sat Jan 16, 2016 1:17 pm
- Forum: Secondary Level
- Topic: An Easy Geometry Problem
- Replies: 2
- Views: 3426
An Easy Geometry Problem
A triangular pyramid has edge of length $\sqrt{2}$, the
volume is $\frac{a}{b}$ where a, b are co-primes. b-a = ?
volume is $\frac{a}{b}$ where a, b are co-primes. b-a = ?
- Sat Jan 09, 2016 11:20 am
- Forum: Number Theory
- Topic: Number Theory
- Replies: 0
- Views: 2082
Number Theory
The number of factor(s) of $(n+1)!$ is double than the number of factor(s) of $n!$. Now, find all non-prime values of $(n+1)$.
- Fri Sep 04, 2015 9:13 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2000/1
- Replies: 6
- Views: 5841
Re: IMO 2000/1
As $AB || CD$, we get $\angle EBA = \angle EDC$. Again by Alternate Segment Theorem, $\angle ABM = \angle BDM$. So, $\angle EBA = \angle ABM$. Similarly we get $\angle EAB = \angle BAM$. So,we get $\triangle EBA \cong \triangle ABN$ by the ASA theorem. Then, we get $EN \perp AB.$ Now, let the segme...
- Thu Aug 27, 2015 11:05 pm
- Forum: National Math Camp
- Topic: Exam 2, Online Number Theory Camp, 2015
- Replies: 24
- Views: 22684
Re: Exam 2, Online Number Theory Camp, 2015
৯ টা বাজে না ক্যারে?