24Looks like no one is trying ..
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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32 Answers
1Q5) LHS=(1-b/a)(1-c/a)+(1-a/b)(1-c/b)+(1-a/c)(1-b/c)
=[(a-c)(a-b)/a]+[(b-a)(b-c)/b]+[(c-a)(c-b)/c]
=[-a2bc-ab2c-abc2+a2b2+b2c2+a2c2]/abc
={[-a2bc-ab2c-abc2]/abc}+{a2b2+b2c2+a2c2]/abc}
=-(a+b+c) +{a2b2+b2c2+a2c2]/abc}
Now [a2b2+b2c2+a2c2]/abc ≥a+b+c for all a,b,c>0
so -(a+b+c) +{a2b2+b2c2+a2c2]/abc}≥0
i.e. (1-b/a)(1-c/a)+(1-a/b)(1-c/b)+(1-a/c)(1-b/c) ≥0