Here are two good questions from Mathlinks , maybe you should give it a try .
1 > Let the positive integers a 1 , a 2 , a 3 , a 4 .... satisfy this eqn . ,
{ 1 / √ a 1 } + { 1 / √ a 2 } + .... { 1 / √ a 100 } = 20 .
Prove that there must two integers which are same .
P . S -------- >> What would have happened if the sum was 19 instead of 20 ???
{ this actually is a hint : ) ; ) }
2 . Prove the following inequality -
3 { a + √ a b + 3√ a b c } ≤ { 8 + 2 √ a ba + b } { 3√a ( a + b ) ( a + b + c ) 6 }
where a , b , c are positive reals .
11For the 1st sum, u simply need an inequality.
\frac{1}{\sqrt{k}}<2(\sqrt{k}-\sqrt{k-1}) for naturals k.
For any set of 100 naturals (denote them by a_i,) we have \sum_{i=1}^{100}\left(\frac{1}{\sqrt{a_i}} \right)<\sum_{x=1}^{100}\left(\frac{1}{\sqrt{x}} \right)
We have from the above stated inequality for the 1st 100 naturals, \sum_{x=1}^{100}\left(\frac{1}{\sqrt{x}} \right)<20 , proving there exists at least 2 naturals in the given set which are equal.
1By Holder's inequality ,
a + √ a b + 3√ a b c ≤ { a + a + a } 1 / 3 { a + √ a b + b } 1 / 3 { a + b + c } 1 / 3
≤ 3 2 / 3 { a + √ a b + b } 1 / 3 { 2a + b } 1 / 3 { 3√a (a + b ) ( a + b + c )6 }
It remains to show that ,
3 5 / 2 { a + √ a b + b } 1 / 3 { 2a + b } 1 / 3 ≤ 8 + 2 √ a ba + b
Or equivalently ,
3 5 { 2 + 2 √ a ba + b } ≤ { 8 + 2 √ a ba + b } 3
Let x = 2 √ a ba + b .
Using AM - GM ,
3 5 { 2 + x } = 3 3 [ 3 . 3 . { 2 + x } ] ≤ 3 3 { 3 + 3 + 2 + x3 } 3 = { 8 + x } 3
You could look up Wikipedia for a slightly modified inequality of this one , but I forgot its name .
