The sum of several positive real numbers equal 10, and the sum of squares of these numbers is not less than 20. Prove that the sum of cubes is at least 40
@eliassaab @oldrin.bataku
several positive worst case all of them = 1 so 10 number ( if we dint count 0 case )
hmm there is much worst case since its real not integeral hmm
ok let it be \(\large a_0+a_1+a_2+...+a_n=10\) \(\large a_0^2+a_1^2+a_2^2+...+a_n^2\ge 20\)
wanna prove \(\large a_0^3+a_1^3+a_2^3+...+a_n^3\ge 40\)
0.001 + 0.0001 + 0.1 + sqrt(2) + ... = 10
it need not be integers right
i already count that case
okay
i was thinking of its enough to prove \(\large a_na_n^2\ge 2a_n+a^2\) it work only for \(\large a_n > 1 \) for <1 or =1 there is a problem so what to do ?
humm so all the several numbers cant be all <=1 at the same time :-|
if there is such away its like devides the numbers into two groups a_n >1 and a_n <=1 and say 1_ not all of them are 1 2_ not all of them are <1 (couse both of them wont satisfy : the sum of squares of these numbers is not less than 20. ) so there would be two cases :- case 1:- \(\large a_0+a_1+a_2+...+a_n=10\) s.t a's >1 case 2 :- \(\large a_0+a_1+a_2+...+a_n+ g_0+g_1+g_2+...+g_n=10\) s.t a's>1 , and g's <1
case 1 easy to prove :D \( \large a_0+a_1+a_2+...+a_n=10...*\) \(\large a_0^2+a_1^2+a_2^2+...+a_n^2\ge 20....**\) \(\large a_0a_0^2+a_1a_1^2+a_2a_2^2+...+a_na_n^2 ...****\) or multiply * with ** ur gonna have lower bound of 200 ( it would be complicated ) so for *** we need to show \(\large a_0a^2\ge 2a_n+a_n^2\) ( prove it bye induction for \(a_n \ge 2\) it will work and since its true sentence ) then \(\large a_0a_0^2+a_1a_1^2+a_2a_2^2+...+a_na_n^2 ...\) \(\large 2(a_0+a_1+a_2+...+a_n)+(a_0^2+a_1^2+a_2^2+...+a_n^2)\ge 2(10)+20=40 \) so we proved hat \(a_0^3+a_1^3+a_2^3+...+a_n^3\ge \) for a'2 >=2 xD more cases incomming
@Mashy take a look if u have time :D
O.o
nvm lolz at least i got an idea :3
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