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7 Oct 2008 20:05:39 IST

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For the Inequality Enthusiasts - II (easy)

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If a,b and c are non-negative reals such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2$ , prove that $abc \le \frac{1}{8}$

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Decoder

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8 Oct 2008 13:03:16 IST

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given

so by A.M-G.M..

so by H.M-G.M..

Decoder

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8 Oct 2008 13:06:52 IST

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oh!..tht's h.m-a.m in the last step..

trying the first one as well..

Hari Shankar

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8 Oct 2008 13:14:10 IST

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Wait, before you go to the other one why dont you fix the problems in this solution:

(a) Your are not given that abc<= 1/8. You have to prove that

(b) Even so, if abc<=1/8 AM-GM is not guilty of stating that a+b+c>=3/2

Lots of work to do man!

Decoder

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8 Oct 2008 13:19:22 IST

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oops!! ..did the other way..round..

i think..there is no. problem in c) ..if there is plz. clear..

in b)..can't we write..(a+b+c)^3 >= 27abc..so derived. from there..

plz. clear..

Anand Hegde

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9 Oct 2008 21:32:38 IST

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edited

sutanoy dasgupta

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10 Oct 2008 22:47:17 IST

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taking l.c.m. of the given,expression and taking the denominator to
the R.H.S.,we get
1>=2abc+ab+bc+ca.
now,ab+bc+ca>=3{(abc)^2/3}
let abc=x^3.
so,1>=2x^3 +3x^2.
2x^3 -x^2 +4x^2-2x+2x-1<=0.
or,(x-0.5)(2.x^2+4x+2)<=0.
but 2.x^2+4x+2>=0 for all x.
so, x<=0.5.
x^3<=1/8
abc<=1/8.

Hari Shankar

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11 Oct 2008 09:34:19 IST

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nice solution ronty!

This one had been posted on mathlinks actually. The solution I gave was:

$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2 \Rightarrow \frac{1}{1+c} \ge \left(1-\frac{1}{1+a} \right) +\left(1-\frac{1}{1+b} \right) \\ \\<br/>= \frac{a}{1+a} + \frac{b}{1+b} \ge 2 \sqrt {\frac{ab}{(1+a)(1+b}}$

Likewise we have

$\frac{1}{1+b} \ge 2 \sqrt {\frac{ac}{(1+a)(1+c}}$ and $\frac{1}{1+a} \ge 2 \sqrt {\frac{bc}{(1+b)(1+c}}$

Multiplying the three inequalities we get

$1 \ge 8 abc$ or $\frac{1}{8} \ge abc$

I am happy this forum is coming alive like this with variety of solutions to problems.

Dipanjan

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11 Oct 2008 11:40:12 IST

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I Have another solution.

Hari Shankar

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11 Oct 2008 12:03:34 IST

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Wow this is really getting better and better. Way to go guys!!

Hari Shankar

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11 Oct 2008 12:06:17 IST

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And by the way, while your hols were on I had posted this one http://goiit.com/posts/list/algebra-for-the-inequality-enthusiasts-i-84890.htm

I will be supremely happy if you guys crack this one!

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