341Q5 - check post No.10 in this thread
Didnt want to make the previous thread tooo long..http://targetiit.com/iit-jee-forum/posts/inequalities-sikhna-chahte-ho-to-sikho-varna-koi-b-11546.html
so new questions in new thread...
This is season 2 of "Inequalities Sikhna chahte ho to sikho varna koi baat nain :P" [3][3]
Q1 Prove that 1.3.5.7..(2n-1)2.4.6...(2n)>1n+1
If n is a positive integer.. Prove that
Q2 2n(n!)≤(n+1)n
Q3 (1-1n)n<(1-1n+1)n+1
Q4 (1+1n)n+1>(1+1n+1)n+2
Q5 For all a,b,c >0, prove that (1-\frac {b}{a})(1-\frac {c}{a})+(1-\frac {a}{b})(1-\frac {c}{b})+(1-\frac {a}{c})(1-\frac {b}{c})\geq 0
Q6 Prove that \frac {\sqrt{1} +\sqrt {1/2} +........+\sqrt {1/n}}{\sqrt{n}}\leq (2n-1)^{1/4}
where n is a positive integer
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UP 0 DOWN 0 4 32
32 Answers
1\frac{3}{4}>\frac{2}{3}\\\\ \frac{5}{6}>\frac{3}{4}\\\\ \frac{7}{8}>\frac{4}{5}\\\\ . \\.\\ .\\\\ \frac{2n-1}{2n}>\frac{n}{n+1}\\\\ multiplying\ all\ inequation\\\\ \frac{3.5.7.9...(2n-1)}{4.6.8...2n}>\frac{2}{(n+1)}\\\\ \frac{1.3.5.7.9...(2n-1)}{2.4.6.8...2n}>\frac{1}{(n+1)}\\\\
11http://www.goiit.com/posts/list/algebra-inequality-88595.htm for no 6.
Just conc on prophet sir's soln or mine, leave out the following trash !
4Mistake in Honey's solution
Ist step :
[(a-c)(a-b)/a2]+[(b-a)(b-c)/b2]+[(c-a)(c-b)/c2]
24Q3,4 corrected now [1]
and Q5 still not solved....can anyone spot mistake in honey's solution ???
1thank god...i was right
Q9 is wrong....generally i do suffer from this prob of lack of confidence....after my post no1 said anything abt Q9....so i was worried if i am wrong
4I think now all the 10 probs r solved
In Qs 9 the sign has to be ≥ instead of > (Check for n=1)
49 ) We want to prove the inequality: ({n+1}/2)n({2n+1}/3)n > (n!)2
Notice that the following equality holds for each positive integer n (it can be proved by induction on n):
12+22+...+n2={n(n+1)(2n+1)/{6}
Therefore considering the left part of our desired inequality,we have:
({n+1}/2)n ({2n+1}/3)n = {(n+1)(2n+1)/6)n =({12+22+...n2}/n)n > 12 22 ...n2 = ( n!)2
because of the AM-GM.
24nice work dimensions..
btw this thread needs to be completed quickly..........
410 ) Use 1/m - 1/(m+1) < 1/m2 < 1/(m-1) - 1/m
1/(n+1) -1/(n+2) + 1/(n+2) ...- 1/(n+p+1) < 1/(n+1)2 + ...+ 1/(n+p)2 < 1/n - 1/(n+1) +
1/(n+!) - ... - 1/(n+p)
=> 1/(n+1) - 1/(n+p-1) < 1/(n+1)2 + ...+ 1/(n+p)2 < 1/n - 1/(n+p)
1Q7)
\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n^2-n)}>\frac{n^2-n+1}{n^2}\\\\\\ \frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n^2-n)}>1-(\frac{n-1}{n^2})\\\\\\ as\ n\geq 1,clearly,\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n^2-n)}>1
1Q3)
let\ (1+\frac{1}{n})=x \\\\ so\ we\ have\ to\ prove\\\\ \frac{2x-1}{x}\ge x^{\frac{1}{x}}\\\\ by\ AM-GM\ inequality\\\\ \frac{1+1+1+...+1+x}{x}\ge x^{\frac{1}{x}}\\\\ =>\ \ \frac{2x-1}{x}\ge x^{\frac{1}{x}}
1apply AM>GM on first n natural number's
\frac{1+2+3+...+n}{n}\geq ({1.2.3.4...n})^\frac{1}{n} \\\\ \frac{n+1}{2}\geq (n!)^\frac{1}{n}\\\\2^n(n!)\leq (n+1)^n
24Q7For \ n\geq 1, \ prove \ that \frac {1}{n} +\frac {1}{n+1} +\frac {1} { n+2}+.....+\frac {1}{n^2}>1
Q8 If a,b,x,y be real numbers such that a2+b2=1,x2+y2=1 then prove that ax+by≤1
Q9 If n is a positive integer prove that
(\frac {n+1}{2})^n (\frac {2n+1}{3} )^n >(n!)^2
Q10 I n and p are positive integers and n≥1 ,p≥1 prove that \frac {1}{n+1} -\frac {1}{n+p+1}<\frac {1}{(n+1)^2}+\frac {1}{(n+2)^2}+...+\frac {1}{(n+p)^2}<\frac {1}{n} -\frac {1}{n+p}
1tchebychefs inequality is a very common type of inequality....
just do a wikipedia search for it....
or u can search eureka's forum for it....i once found some very good proofs there.....
1yes sir the
question must be this-
instead of n/n+1 it should be
(n+1)/n
341but arshad if http://targetiit.com/iit-jee-forum/posts/prove-it-good-one-11869.html is true then in Q9 the direction of the inequality is wrong
1for Q no.6
just apply tchebychefs inequality twice
and for q no.9
just apply a.m.>g.m (for squares of number till n)
as simple as dat
1oh yeah prophet sir,...with Trigonometry the q8 was just a single step problem...
341busy with sth. i had posted something similar some time ago. To make it easy to understand, its called the ε-δ challenge.
I'll post after some time.
341For q5) you can use an inequality thats easy to remember: http://en.wikipedia.org/wiki/Schur%27s_inequality
1Q8. x2+y2 =1, a2+b2=1,
Multiplying both these,
a2x2+b2x2+a2y2+b2y2=1.
a2x2+b2y2=1-a2y2-b2x2
(ax+by)2= 1-(ay-bx)2
ax+by=√1-(ay-bx)2
thus,ax+by≤1.
1the inequality is wrong i think.......no positive integer satisfies the inequality
1Q6
SOLN:
the series can be written as √1/n
=> (√1/n)/√n ≤ (2n - 1)1/4
= 1/n4 ≤ 2n - 1 -----------------------(taking the 4th power on both sides)
= 1 ≤ 2n5 - n4 -----------------(multiplying n4 on both sides)
NOTE : there will be no change in sign since n is positive integer
HENCE PROVED
(try putting any positive Integer value of n..........it will definately satisfy the inequality)
1Q4) (1 + 1/n)n+1>(1 + 1/n+2)n+1
= 1+ 1/n > 1 + 1/n+2 -------------------(Taking the (n+1)th root)
= 1/n > 1/n+2 ------------------------(subtracting 1 from both sides)
simple sa logic lagao..............kyunki RHS ka denominator bada hai isiliye RHS chhota fraction hai
aur agar detail mein jaana hai toh iss sum ko aur lamba kheech sakte ho