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Inequalities Marathon
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peine
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#281
my solution is also with a lot of calculs, then you can post your solution. Smile
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PostPosted: Sat Oct 31, 2009 3:55 am  Back to top 
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hasan4444
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#282
OK sorry "peine" but this is a Pre-Olympiad marathon Mr. Green and calculus is not really welcomed Smile

Looking forward for your next post
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PostPosted: Sat Oct 31, 2009 5:03 am  Back to top 
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#283
solution to problem 132

2a + 3b + 3c = abc \implies a(bc - 2) = 3(b + c)

\implies a + b + c = \frac {3(b + c)}{bc - 2} + b + c \geq \frac {6\sqrt {bc}}{bc - 2} + 2\sqrt {bc}
Let \sqrt {bc} = k and sure k^2 > 2 .Then we're looking to minimize the function
f(k) = \frac {6k}{k^2 - 2} + 2k = \frac {2k^3 + 2k}{k^2 - 2} \ ; \implies f'(k) = \frac {10k^4 - 6k^2 - 4}{(k^2 - 2)^2} = 0

\iff 10k^4 - 6k^2 - 4 = 0 \iff (k^2 - 1)(5k^2 + 2) = 0
Function f is increasing in (1,\infty) so in the interval (2,\infty) the minimum occur for k^2 \to 2

We do not need to find \lim_{k^2 \to 2} {\frac {2k^3 + 2k}{k^2 - 2}} , we just use b = c = \sqrt 2 ,since b+c \geq 2k ,and because equality occur for b=c

For b= c = \sqrt 2 ,we have a+ 2\sqrt 2 = 2a \implies a = 2\sqrt 2

Thus the searched minimum is a + b + c = 4\sqrt 2


I hope I have no mistakes in calculations ,but anyway this is the idea.



Problem 133.Let a,b,c,d be four positive real number with sum 4 .Prove that
\frac {a + 1}{b^2 + 1} + \frac {b + 1}{c^2 + 1} + \frac {c + 1}{d^2 + 1} + \frac {d + 1}{a^2 + 1} \geq 4
\[
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PostPosted: Sat Oct 31, 2009 6:02 am  Back to top 
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#284
solution to 133


First notice that:

\frac {a + 1}{b^2 + 1} = \frac {(a + 1)(b^2 + 1)}{b^2 + 1} - \frac {b^2(a + 1)}{b^2 + 1} = a + 1 - \frac {b^2(a + 1)}{b^2 + 1...

since

\frac {b^2(a + 1)}{b^2 + 1} \leq \frac {b^2(a + 1)}{2b} \implies - \frac {b^2(a + 1)}{b^2 + 1} \ge - \frac {b^2(a + 1)}{2b}

and we get:

\sum_{cyc} \frac {a + 1}{b^2 + 1} \ge \sum_{cyc} a + 1 - \frac {ab}{2} - \frac {b}{2} \ge 4


Problem 134.(edited)

If a,b,c,d are positive numbers prove that:
\sum_{cyc} \frac{a}{b+2c+3d} \ge \frac{2}{3}
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Last edited by Obel1x on Sat Oct 31, 2009 10:49 am; edited 1 time in total 
PostPosted: Sat Oct 31, 2009 9:30 am  Back to top 
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peine
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#285
I don't see the problem 134, anyway this is another solution to problem 133, I wan a post it before that but I had a problem in conexion
another solution to problem 133
we have the function , f(x)=\frac{1}{x^2+1} is convex and decreasing in \mathbb{r} then from Jensen,
\frac{a}{4}f(b)+\frac{b}{4}f(c)+\frac{c}{4}f(d)+\frac{d}{4}f(a)\geq f\left(\frac{ab+bc+cd+da}{4}\right) and
f(a)+f(b)+f(c)+f(d)\geq 4f\left(\frac{a+b+c+d}{4}\right)=4f(1) then:
L.H.S \geq 4f(1)+4f\left(\frac{ab+bc+cd+da}{4}\right)
and because that ab+bc+cd+da\leq4 and f is decreasing we get:
L.H.S \geq 8f(1)=4

to Hassan:
I'm sorry for the last inequality that I post
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PostPosted: Sat Oct 31, 2009 9:47 am  Back to top 
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Abdek
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#286
Solution to problem 134:

By cauchy shwraz inequality we have :

\sum_{cyc}\frac {a}{b + 2c + 3d}.\sum_{cyc}a(b + 2c + 3d) \ge (a + b + c+d)^2

Which is equivalent to :

LHS \ge \frac {(a + b + c + d)^2}{4(ab + bc + cd + da + ac + bd)}

And hence it remains to prove that:

3(a + b + c + d)^2 \ge 8(ab + bc + cd + da + ac + bd)

which is equivalent to

(a - b)^2 + (b - c)^2 + (c - d)^2 + (d - a)^2 + (a - c)^2 + (b - d)^2 \ge 0

which is true .

PostPosted: Sat Oct 31, 2009 1:07 pm  Back to top 
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Abdek
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#287
Problem :135

Let a,b,c>0. Prove that:

\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}+abc \le \frac{a^7}{b^2c^2}+\frac{b^7}{a^2c^2}+\frac{c^7}{a^2b^2}+\frac{1}{a^2b^2...

PostPosted: Sat Oct 31, 2009 1:18 pm  Back to top 
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peine
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#288
solution to problem 135
the inequality is equivalent as:
a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 \leq a^8 + b^8 + c^8 + 1
by AM-GM two times, we have,
R.H.S = \sum \left{3\times\frac {1}{3}a^8 + 1} \geq \frac {4}{3}(a^6 + b^6 + c^6) \geq a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 =...
equality holds when a = b = c = 1

Problem 136: (Mohamed El-Alami)
let a,b,c be positive real numbers, prove that:
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} \geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}+\frac{3}{2}
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Last edited by peine on Sun Nov 01, 2009 2:56 am; edited 1 time in total 
PostPosted: Sun Nov 01, 2009 2:10 am  Back to top 
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#289
peine wrote:
solution to problem 135
the inequality is equivalent as:
a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 \leq a^8 + b^8 + c^8 + 1
by AM-GM two times, we have,
R.H.S = \sum \left{3\times\frac {1}{3}a^8 + 1} \geq \frac {4}{3}(a^6 + b^6 + c^6) \geq a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 =...
equality holds when a = b = c = 1

Problem 136: (Mohamed El-Alami)
let a,b,c be positive real numbers, prove that:
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} \geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}


Look at the first problem in this marathon Wink

PostPosted: Sun Nov 01, 2009 2:29 am  Back to top 
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peine
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#290
I'm sorry It was a mistake, it's edited now Smile
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PostPosted: Sun Nov 01, 2009 2:58 am  Back to top 
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geniusbliss
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#291
peine wrote:
solution to problem 135
the inequality is equivalent as:
a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 \leq a^8 + b^8 + c^8 + 1
by AM-GM two times, we have,
R.H.S = \sum \left{3\times\frac {1}{3}a^8 + 1} \geq \frac {4}{3}(a^6 + b^6 + c^6) \geq a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 =...
equality holds when a = b = c = 1

Problem 136: (Mohamed El-Alami)
let a,b,c be positive real numbers, prove that:
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} \geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b} + \frac {3}{2}

this is equivalent to the japan tst 2004(note which was later in 2009 copied in AMTI Inter Embarassed )
i shall post a new inequality-
Problem 137
Prove that for all nonegative reals a,b,c we have ,
a^3 + b^3 + c^3 - 3abc\ge4(a - b)(b - c)(c - a)
see this link for the problem posted by peine: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1520533121&t=25780
or this one - http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=105567310&t=209772
Last edited by geniusbliss on Sun Nov 01, 2009 4:58 am; edited 1 time in total 
PostPosted: Sun Nov 01, 2009 4:27 am  Back to top 
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#292
Solution to problem 136

With AM-GM we have :
\frac {1}{2}(\frac {a}{b} + \frac {b}{c} + \frac {c}{a})\geq \frac {3}{2}
So we have only to prove :
\frac {1}{2}(\frac {a}{b} + \frac {b}{c} + \frac {c}{a})\geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}
We consider :
\frac {a}{b} = x ; \frac {b}{c} = y;\frac {c}{a} = z
L'inequality becomes:
\frac {1}{2}(x + y + z)\geq \frac {xy}{x + y} + \frac {yz}{y + z} + \frac {zx}{z + x}
and ,
\frac {xy}{x + y}\leq \frac {x + y}{4}
because
(x + y)^{2}\geq 4xy
the same thing for :
\frac {yz}{z + y}\leq \frac {z + y}{4}
and
\frac {zx}{x + z}\leq \frac {x + z}{4}
And finally :
\frac {1}{2}(x + y + z)\geq \frac {xy}{x + y} + \frac {yz}{y + z} + \frac {zx}{z + x}

Last edited by b.s.o on Sun Nov 01, 2009 1:24 pm; edited 1 time in total 
PostPosted: Sun Nov 01, 2009 4:41 am  Back to top 
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#293
geniusbliss wrote:
peine wrote:
solution to problem 135
the inequality is equivalent as:
a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 \leq a^8 + b^8 + c^8 + 1
by AM-GM two times, we have,
R.H.S = \sum \left{3\times\frac {1}{3}a^8 + 1} \geq \frac {4}{3}(a^6 + b^6 + c^6) \geq a^3b^3 + b^3c^3 + c^3a^3 + a^3b^3c^3 =...
equality holds when a = b = c = 1

Problem 136: (Mohamed El-Alami)
let a,b,c be positive real numbers, prove that:
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} \geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b} + \frac {3}{2}

this is equivalent to the japan tst 2004(note which was later in 2009 copied in AMTI Inter Embarassed )
i shall post a new inequality-
Problem 136
Prove that for all nonegative reals a,b,c we have ,
a^3 + b^3 + c^3 - 3abc\ge4(a - b)(b - c)(c - a)
see this link for the problem posted by peine: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1520533121&t=25780
or this one - http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=105567310&t=209772


Sorry i didn't see the new problem

PostPosted: Sun Nov 01, 2009 4:54 am  Back to top 
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hasan4444
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#294
geniusbliss wrote:

this is equivalent to the japan tst 2004(note which was later in 2009 copied in AMTI Inter Embarassed )
i shall post a new inequality-


I'm getting tired again how many time should I repeat this for you "geniusbliss" enough is enough again you didn't put a solution and your excuse that it is equivalent to whatever, then so do you really think that all the inequalities so far are own 100% really?! I really wish to block you from posting here but uncooperative admins are not helping me.
Yes and don't now post a new post for another silly thing OK. You have a solution or a problem or a creative post do it otherwise stay back.

However, this is the pending problem:
Problem 136
Prove that for all nonegative reals a,b,c we have ,
a^3 + b^3 + c^3 - 3abc\ge4(a - b)(b - c)(c - a)

Have Fun Mathematicians!!!
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PostPosted: Sun Nov 01, 2009 4:58 am  Back to top 
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#295
b.s.o wrote:
\frac {a}{b} = x ; \frac {b}{c} = y;\frac {c}{a} = z
L'inequality becomes:
\frac {1}{2}(x + y + z)\geq \frac {xy}{x + y} + \frac {yz}{y + z} + \frac {zx}{z + x}


I think you're wrong in this step .

b.s.o wrote:

\frac {1}{2}(\frac {a}{b} + \frac {b}{c} + \frac {c}{a})\geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}


This is true only if a,b,c are sidelenghts of a triangle .

PostPosted: Sun Nov 01, 2009 5:00 am  Back to top 
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geniusbliss
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#296

@hasan i am very sorry
ok sorry guess nobody can be more dumb than me -
anyway
why it is equivalent-
the japan tst is -
for a + b + c = 1 and a,b,c positive reals prove that
\sum_{cyc}\frac {1 + a}{1 - a} \le \sum_{cyc}\frac {a}{b}
write 1 + a = 2a + (b + c) and 1 - a = b + c,
so the LHS becomes \sum_{cyc}\frac {2a}{b + c} + 3 now divide the LHS and RHS by 2 and we get -
\sum_{cyc}\frac {a}{b} \ge \frac {3}{2} + \sum_{cyc}\frac {a}{b + c}
since the inequality posted by peine is homogenous we assume that a + b + c = 1
so the inequalities are equivalent.
solution to 136

we have to prove that ,
\sum_{cyc} (\frac {a}{b} - \frac {a}{b + c}) \ge \frac {3}{2}
or,
\sum_{cyc}\frac {ac}{b(b + c)} \ge \frac {3}{2}
by Cauchy,
\sum_{cyc}\frac {ac}{b(b + c)} \ge \sum_{cyc}\frac {(ac)^2}{acb(b + c)} \ge \frac {(ab + bc + ca)^2}{2abc} \ge \frac {3}{2} and the last inequality is by AM-GM for positive reals ab,bc,ca

solution to 137

(not mine)
let c = min {a,b,c} ,
then we substitute {a,b,c} to {a - c,b - c,c - c} since a,b,c are non-negative reals,
we observe that the LHS of the inequality gets decreased (from (a + b + c) to (a + b - 2c)) as one of the factors and the other factor is unaltered
Also that the RHS remains the same 4(a - b)(b - c)(c - a) so we are done if we prove this new inequality instead of the earlier one.
after the substituion the inequality becomes -
a^3 + b^3\ge 4ab(b - a)
for this: By AM- GM we have \frac {b^2}{4} \ge a(b - a)
therefore,
4ab(b - a) \le 4b.\frac {b^2}{4} = b^3 \le a^3 + b^3 (note that the last inequality has an equality case as a,b,c are non-negative.)
thus proved,with equality for a = b = c

@hasan
i am very sorry dude.
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PostPosted: Sun Nov 01, 2009 5:20 am  Back to top 
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hasan4444
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#297
Problem 138
Gabriel Dospinescu, Marian Tetiva
Problem 138: Let x,y,z >0 such that
x+y+z=xyz
Prove that
(x-1)(y-1)(z-1) \le 6 \sqrt{3} -10
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PostPosted: Mon Nov 02, 2009 5:11 am  Back to top 
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peine
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#298
very nice problem, hope someone find a solution nicer than mine,
Solution to problem 138:

Let P = (x - 1)(y - 1)(z - 1) and let find P_{max}
we have xyz = x + y + z \Leftrightarrow x = \frac {y + z}{yz - 1}
then:
P = \left(\frac {y + z}{yz - 1} - 1\right)(y - 1)(z - 1)
\Leftrightarrow (yz - 1)P = (y - 1)(z - 1)(2 - (y - 1)(z - 1))
we have the function g(a) = a(2 - a) is increasing in [ - \infty,1] and we have (y - 1)(z - 1)\leq 1 and (x - 1)(y - 1)\leq (\sqrt {yz} - 1)^2 then (yz - 1)P \leq (\sqrt {yz} - 1)^2(2 - (\sqrt {yz} - 1)^2)
\Leftrightarrow P\leq \frac {2t - t^3}{t + 2} = f(t) with t = \sqrt {yz} - 1 , t\geq 0 (easy to prove)
f'(t) = \frac {4 - 2t^3 - 6t^2}{(t + 2)^2}
and from that we can easly verify that f(t) is maximal in [0, + \infty] when t = \sqrt {3} - 1
from where we deduct that P_{max} = 6\sqrt {3} - 10 equality holds when x = y = z = \sqrt {3}

Problem 139: (Mohamed El-Alami)
Let a,b,c be positive real numbers, find the greatest constant k such as:
\frac {a^3 + b^3 + c^3}{3abc} + \frac {k(ab + bc + ca)}{a^2 + b^2 + c^2} \geq k + 1
P.S: I found this result today, hope that was really my own result.
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PostPosted: Mon Nov 02, 2009 2:40 pm  Back to top 
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#299
Solution to problem 139:

Nice problem my friend

your inequality is equivalent to :

\frac {a^3 + b^3 + c^3}{3abc} - 1 + \frac {k(ab + bc + ca - a^2 - b^2 - c^2)}{(a^2 + b^2 + c^2)} \ge 0


by the two flowing identities a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)\left((a - b)^2 + (b - c)^2 + (c - a)^2\right) and
a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left((a - b)^2 + (b - c)^2 + (c - a)^2\right)
The inequality can be written as :
\left((a - b)^2 + (b - c)^2 + (c - a)^2\right)\left(\frac {a + b + c}{3abc} - \frac {k}{a^2 + b^2 + c^2}\right) \ge 0

\Leftrightarrow (a^2 + b^2 + c^2)(a + b + c) \ge 3.k.abc

which is 3 by AM_GM Smile
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PostPosted: Tue Nov 03, 2009 4:47 am  Back to top 
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#300
Problem 140:

For a,b,c>0 show that :

\frac{a^9}{bc}+\frac{b^9}{ca}+\frac{c^9}{ab}+\frac{2}{abc} \ge a^5+b^5+c^5+2
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PostPosted: Tue Nov 03, 2009 5:09 am  Back to top 
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