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Inequalities Marathon
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alex2008
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#21
Solution to problem 10
Just observe that a^3+b^3+c^3+d^3\le 2(a^2+b^2+c^2+d^2)=8 because a,b,c,d\le 2


I won't post other problem because the solution to problem 8 is not correct .
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PostPosted: Thu Sep 10, 2009 12:35 am  Back to top 
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aadil
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#22
sorry hasan Embarassed
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PostPosted: Thu Sep 10, 2009 3:57 am  Back to top 
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enndb0x
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#23
solution to problem 8


Let b=a+x ,c=b+y =a+x+y ,sure x,y\geq 0

Inequality becomes
(2a+x)(x+y+2a)^2 -6a(a+x)(a+x+y) \geq 0

But

(2a+x)(x+y+2a)^2 -6a(a+x)(a+x+y) =2a^3 +2a^2 y +2axy +2ay^2 +x^3 +2x^2 y +xy^2 ,which is clearly positive



Problem 11

Let a,b,c be positive real numbers such that abc=1 .Prove that
\sum_{cyc}{\frac{1}{a^2 +2b^2 +3} \leq \frac{1}{2}

PostPosted: Thu Sep 10, 2009 4:44 am  Back to top 
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alex2008
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#24
Solution to problem 11
Using AM-GM we have :

LHS= \sum_{cyc}\frac{1}{(a^2+b^2)+(b^2+1)+2}\le \sum_{cyc} \frac{1}{2ab+2b+2} =\frac{1}{2}\sum_{cyc} \frac{1}{ab+b+1}=\frac{1}{2}

because \frac{1}{bc+c+1}=\frac{1}{bc+c+abc}=\frac{1}{c}\cdot \frac{1}{ab+b+1}=\frac{ab}{ab+b+1}

and \frac{1}{ca+a+1}=\frac{1}{\frac{1}{b}+a+1}=\frac{b}{ab+b+1}

so \sum_{cyc}\frac{1}{ab+b+1}=\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}=1


Problem 12. Let a,b,c>0 such that a+b+c=1 . Prove that:

\frac{1+a+b}{2+c}+\frac{1+b+c}{2+a}+\frac{1+c+a}{2+b}\ge \frac{15}{7}
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PostPosted: Thu Sep 10, 2009 12:44 pm  Back to top 
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Dimitris X
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#25
solution to problem 12

\sum\frac{1+a+b}{2+c}+1 \ge \frac{15}{7}+3 \Longleftrightarrow \sum\frac{3+(a+b+c)}{2+c} \ge \frac{36}{7} \Longleftrightarrow...
But \sum\frac{2^2}{2+c} \ge \frac{(2+2+2)^2}{2+2+2+a+b+c}=\frac{36}{7}


PROBLEM 13
Let a,b,c be positive so that abc=1

\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \le 1

Remark

This was IMO 2000 problem but its solution elementary and attackable for a ''pre-olympiad level'' student Smile

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PostPosted: Thu Sep 10, 2009 1:47 pm  Back to top 
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enndb0x
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#26
solution to problem 13


\left(a - 1 + \frac {1}{b} \right) \left(b - 1 + \frac {1}{c} \right) \left(c - 1 + \frac {1}{a} \right) \leq 1

Substitute a = \frac {x}{y} ,b = \frac {y}{z} and c = \frac {z}{x}

Inequality is equivalent with

\left(\frac {x}{y} - 1 + \frac {z}{y} \right)\left(\frac {y}{z} - 1 + \frac {x}{z} \right) \left(\frac {z}{x} - 1 + \frac {y}...

\iff (x + z - y)(y - z + x)(z - x + y) \leq xyz

WLOG ,Let x > y > z , then x + z > y ,x + y > z .If y + z < x ,then we are done because

(x + z - y)(y - z + x)(z - x + y) \leq 0 and xyz \geq 0

Otherwise if y + z > x , then x,y,z are side lengths of a triangle ,and then we can make the substitution

x = m + n ,y = n + t and z = t + m

Inequality is equivalent with

8mnt \leq (m + n)(n + t)(t + m) ,this is true by AM-GM

m + n \geq 2\sqrt {mn} ,n + t \geq 2 \sqrt {nt} and t + m \geq 2 \sqrt {tm} ,multiply and we're done.

another solution to problem 12


Let a\geq b \geq c then by Chebyshev's inequality we have

LHS \geq \frac {1}{3} \left(1 + 1 + 1 + 2(a + b + c)\right) \sum_{cyc}{\frac {1}{2 + a}} = \frac {5}{3} \sum_{cyc}{\frac {1}{...

By Titu's Lemma \sum_{cyc}{\frac {1}{2 + a}\geq \frac {9}{7} ,then LHS \geq \frac {15}{7}

Problem 14

Let a,b,c > 0 and a + b + c = abc. Prove that:
\frac {1}{\sqrt {a^2 + 1}} + \frac {1}{\sqrt {b^2 + 1}} + \frac {1}{\sqrt {c^2 + 1}} \le \frac {3}{2}

PostPosted: Thu Sep 10, 2009 4:05 pm  Back to top 
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geniusbliss
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#27
is this even correct-
Problem10
let a=2x,b=2y,c=2z,d=2w
ob x,y,z,w \le 1 so substitute in the LHS and we get \sum_{cyc}x^3 \le \sum_{cyc}x^2 which is obvious.

I have a doubt dont u guys think it is wrong to just post one solution for a problem?? i think the marathon is going to fast if atleast 5 prrofs are posted for every problem it would be more interesting and nice to see new ideas from different people or most ppl are just stuck with some sum and without solving go to the next jus cuz someone good like alex,dimitris etc.. has posted a solution and a new problem!!
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PostPosted: Thu Sep 10, 2009 11:56 pm  Back to top 
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Dimitris X
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#28
solution to problem 14

Setting a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z} the condition becomes xy+yz+zx=1,and the inequality:
\sum\frac{x}{\sqrt{x^2+1}} \le \frac{3}{2}.
But \sum\frac{x}{\sqrt{x^2+1}}=\sum\frac{x}{\sqrt{x^2+xy+xz+zy}}=\sum\sqrt{\frac{x}{x+y}\frac{x}{x+z}}
But \sqrt{\frac{x}{x+y}\frac{x}{x+z}}\le \frac{\frac{x}{x+y}+\frac{x}{x+z}}{2}
So \sum\sqrt{\frac{x}{x+y}\frac{x}{x+z}} \le \frac{\frac{x}{x+y}+\frac{y}{x+y}+\frac{x}{z+x}+\frac{z}{z+x}+\frac{y}{y+z}+\frac{z...


PROBLEM 15


If a,b,c \in\mathbb R and a^2+b^2+c^2=3.Find the minimum value of A=ab+bc+ca-3(a+b+c).


geniusbliss

I think it ok if someone that has a nice solution for a problem even (if another solution exist) to post it.BUT i think that the ''spirit'' of the marathon is not to wait 10 different solutions for every problem.Of course it is nice to post different solution someone but we cant wait to find 5 different solutions to continue.....

NO OFFENCE MY DEAR FRIEND
Dimitris

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PostPosted: Fri Sep 11, 2009 3:07 am  Back to top 
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#29
A very cute solution to 14

Substitute a=\tan x,b=\tan y and c=\tan z where x+y+z=\pi
And we are left to prove
\cos x+\cos y+\cos z\le\frac{3}{2}
Which i think is very well known..

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PostPosted: Fri Sep 11, 2009 3:37 am  Back to top 
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#30
for geniusbliss
Another solution to problem 1
Bashing out it gives a^4c^2 + b^4a^2 + c^4b^2 + a^3b^3 + b^3c^3 + a^3c^3 \geq abc(ab^2 + bc^2 + ca^2 + 3abc) which is true because AM-GM gives :

a^3b^3 + b^3c^3 + a^3c^3\ge 3a^2b^2c^2

and by Muirhead :

a^4c^2 + b^4a^2 + c^4b^2\ge abc(ab^2++bc^2+ca^2)


Another solutions to problem 1
Observe that the inequality is equivalent with:

\sum_{cyc}\frac{a^2+bc}{a(a+b)}\ge 3

Now use AM-GM:

\sum_{cyc}\frac{a^2+bc}{a(a+b)}\ge 3\sqrt[3]{\frac{\prod (a^2+bc)}{abc\prod (a+b)}}

So it remains to prove:

\prod (a^2+bc)\ge abc\prod (a+b)

Now we prove (a^2+bc)(b^2+ca)\ge ab(c+a)(b+c)\Leftrightarrow a^3+b^3\ge ab^2+a^2b\Leftrightarrow (a+b)(a-b)^2\ge 0

Multiplying the similars we are done
.

Another solution to problem 3
Use Cauchy-Schwartz and Nesbitt :

(a+b+c)\left(\frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right)\ge \left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\...


Another solution to problem 4
Schur gives 1+9abc\ge 4(ab+bc+ca) and use also 3(ab+bc+ca)\le (a+b+c)^2=1 . Suming is done .


Another solution to problem 6
By Cauchy-Schwatz : \frac {4a}{2a^2 + b^2 + c^2} \le \frac {a}{a^2 + b^2} + \frac {a}{a^2 + c^2}

Then we have RHS \le \sum_{cyc} \frac {a + b}{a^2 + b^2} \le \sum_{cyc} \ \frac {2}{a + b} \le \sum_{cyc} \left(\frac {1}{2a} + \frac {1}...


Another solution to problem 8
Let b = xa\ ,\ c = yb = xya \Rightarrow x,y \ge 1

Then:

\ds\frac {(a + b)(a + c)^2}{3} \ge 2abc

\Leftrightarrow (x + 1)(xy + 1)^2 \cdot a^3 \ge 6x^2ya^3

\Leftrightarrow (x + 1)(xy + 1)^2 \ge 6x^2y

\Leftrightarrow (x + 1)(4xy + (xy - 1)^2) \ge 6x^2y

\Leftrightarrow 4xy + (xy - 1)^2\cdot x + (xy - 1)^2 - 2x^2y \ge 0

We have that:

4xy + (xy - 1)^2\cdot x + (xy - 1)^2 - 2x^2y\ge

\ge 4xy + 2(xy - 1)^2 - 2x^2y (\text{ because }x \ge 1)

= 2x^2y^2 + 2 - 2x^2y = 2xy(y - 1) + 2 > 0

done.


but five solutions for each problem ???? that's too much .
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PostPosted: Fri Sep 11, 2009 3:42 am  Back to top 
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geniusbliss
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#31
fine u guys carrry on in your way and me on my own way just dont rebuke if i posted solution for some problem far back.
Dimitris

maybe not 5 or 10 solutions but still..
why are you in such a hurry is the marathon some preparation for some exam coming up or what?? there are no limits in terms of time so why go so fast? either ways you guys are so good that you make me look like i cant solve at your speed and typing this as a petty reason,which isnt true(i mean this is not an excuse u r still gud Razz)
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PostPosted: Fri Sep 11, 2009 3:50 am  Back to top 
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#32
Solution to 15

We know that ab+bc+ac\ge\frac{-(a^{2}+b^{2}+c^{2})}{2}
ab+bc+ac\ge\frac{-3}{2}.Putting this in our inequality we get

A\ge\frac{-3}{2}-3(a+b+c) now squaring both the sides \left(A+\frac{3}{2}\right)^{2}\ge9(a^{2}+b^{2}+c^{2}+2(ab+bc+ac))

Again putting the above values we get \left(A+\frac{3}{2}\right)^{2}\ge0


Tell whether it is correct or not??
If yes then I will post a new problem
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PostPosted: Fri Sep 11, 2009 4:03 am  Back to top 
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#33
solution to problem 15


ab + bc + ca = \frac {(a + b + c)^2 - a^2 - b^2 - c^2 }{2} = \frac {(a + b + c)^2 - 3}{2}

Let a + b + c = x

Then A = \frac {x^2 }{2} - 3x - \frac {3}2}

We consider the second degree fuction f(x) = \frac {x^2}{2} - 3x - \frac {3}{2}

We obtain minimum for f\left(\frac { - b}{2a}\right) = f(3) = - 6

Then A_{min} = - 6 ,it is attained for a=b=c=1


math mechanic
your solution is not correct

geniusbliss
feel free to post a solution anytime you solve a problem

Problem 16 If a,b,c are positive real numbers such that a + b + c = 1.Prove that
\frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\ge \sqrt {\frac {3}{2}}

PostPosted: Fri Sep 11, 2009 4:16 am  Back to top 
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keyree10
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#34
Quote:
Problem 16: If a,b,c are positive real numbers such that a + b + c = 1. Prove that
\frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\ge \sqrt {\frac {3}{2}}


solution
Let f(x) = \frac x {\sqrt {1 - x}}. f"(x) > 0

Therefore, \sum{\frac a {\sqrt {1 - a}} \ge \frac {3s} {\sqrt {1 - s}} ,where s = \frac {a + b + c} 3 = \frac 1 3, by jensen's.

\implies \frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\ge\sqrt {\frac {3}{2}}. Hence proved

P.S.

Just a suggestion : Could we make it a rule to quote the question we are posting a solution to? It'll solve the problem of having to go back a few pages to refer to an earlier question.


PROBLEM 17: If a,b,c are REALS such that a^2 + b^2 + c^2 = 1
Prove that a + b + c - 2abc \le \sqrt2
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PostPosted: Fri Sep 11, 2009 4:44 am  Back to top 
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geniusbliss
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#35
enndb0x wrote:

Problem 16 If a,b,c are positive real numbers such that a + b + c = 1.Prove that
\frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\ge \sqrt {\frac {3}{2}}

solution

By holders' inequality,
(\sum_{cyclic}\frac {a}{(b + c)^{\frac {1}{2}}})(\sum_{cyclic}\frac {a}{(b + c)^{\frac {1}{2}}})(\sum_{cyclic} a(b + c)) \ge ...
thus,
(\sum_{cyclic}\frac {a}{(b + c)^{\frac {1}{2}}})^2 \ge \frac {(a + b + c)^2}{2(ab + bc + ca)} \ge \frac {3(ab + bc + ca)}{2(a...
or,
\frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\ge \sqrt {\frac {3}{2}}
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PostPosted: Fri Sep 11, 2009 5:14 am  Back to top 
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alex2008
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#36
keyree10 wrote:
PROBLEM 17: If a,b,c are REALS such that a^2 + b^2 + c^2 = 1
Prove that a + b + c - 2abc \le \sqrt2


Solution to problem 17
Use Cauchy-Schwartz:

LHS=a(1-2bc)+(b+c)\le \sqrt{(a^2+(b+c)^2)((1-2bc)^2+1)}

So it'll be enough to prove that :

(a^2+(b+c)^2)((1-2bc)^2+1)\le 2\Leftrightarrow (1+2bc)(1-bc+2b^2c^2)\le 1\Leftrightarrow 4b^2c^2\le 1

which is true because 1\ge b^2+c^2\ge 2bc done


Problem 18: Let x,y,z>0 such that xyz=1 . Show that:

x^2+y^2+z^2+x+y+z\ge 2(xy+yz+zx)
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PostPosted: Fri Sep 11, 2009 5:19 am  Back to top 
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#37
Solution to problem 18
To solve the problem of alex, we need Schur and Cauchy inequality as demonstrated as follow
a + b + c \ge 3\sqrt [3]{abc} = \sqrt [3]{(abc)^2} \ge \frac {9abc}{a + b + c} \ge 2(ab + bc + ca) - (a^2 + b^2 + c^2)
Note that we possess the another form of Schur such as
(a^2 + b^2 + c^2)(a + b + c) + 9abc \ge 2(ab + bc + ca)(a + b + c)
Therefore, needless to say, we complete our proof here.


Problem 19. Let a, b, c be positive reals satisfying a^2+b^2+c^2=3. Prove that

(abc)^2(a^3+b^3+c^3) \le 3
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\left(\frac{a_{1}+a_{2}+...+a_{n}}{n} \right) \geq \sqrt[n]{a_1a_{2}a_{3}....a_{n}}

PostPosted: Sat Sep 12, 2009 1:59 am  Back to top 
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#38
great math wrote:
Problem 19. Let a, b, c be positive reals satisfying a^2 + b^2 + c^2 = 3. Prove that
(abc)^2(a^3 + b^3 + c^3) \le 3


Solution to Problem 19
For the sake of convenience, let us introduce the new unknowns u, v, w as follows:
u=a+b+c \\ v=ab+bc+ca\\ w=abc

Now note that u^2-2v=3 and a^3+b^3+c^3=u(u^2-3v)=u\left(\frac{9-u^2}{2}\right).

We are to prove that w^2\left(u\cdot\frac{9-u^2}{2}+3w\right)\le 3.

By AM-GM, we have \sqrt[3]{abc}\le \frac{a+b+c}{3}\implies w\le \frac{u^3}{3^3}.

Hence, it suffices to prove that u^7\cdot\frac{9-u^2}{2}+\frac{u^9}{3^2}\le 3^7.

However, by QM-AM we have \sqrt{\frac{a^2+b^2+c^2}{3}}\ge \frac{a+b+c}{3}\implies u\le 3 which proves the above inequality. \blacksquare

I'm very sorry but I have to leave right now.

Could someone please post a new problem?

Thank you. Smile
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PostPosted: Sat Sep 12, 2009 7:04 am  Back to top 
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#39
Problem 20. Let a, b, c be the lengths of the sides of a triangle. Prove that:
a^2b(a-b)+b^2c(b-c)+c^2a(c-a) \ge 0
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PostPosted: Sat Sep 12, 2009 7:26 am  Back to top 
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#40
hasan4444 wrote:
Problem 20. Let a, b, c be the lengths of the sides of a triangle. Prove that:
a^2b(a - b) + b^2c(b - c) + c^2a(c - a) \ge 0


Solution to problem 20
Use Ravi substitution a=x+y\ ,\ b=y+z\ ,\ c=z+x then the inequality becomes :

\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge x+y+z\ ,

true by Cauchy-Schwartz.

Remark. The inequality is equivalent with :

\frac{1}{2}((a - b)^2(a + b - c)(b + c - a) + (b - c)^2(b + c - a)(a + c - b) + (c - a)^2(a + c - b)(a + b - c))\geq 0


Problem 21.Let x,y,z \in \left[\frac{1}{3},\frac{2} {3}\right] . Show that :

1 \ge \sqrt[3] {xyz}+\frac {2} {3(x+y+z)}

FantasyLover wrote:
great math wrote:
Problem 19. Let a, b, c be positive reals satisfying a^2 + b^2 + c^2 = 3. Prove that
(abc)^2(a^3 + b^3 + c^3) \le 3


Solution to Problem 19
For the sake of convenience, let us introduce the new unknowns u, v, w as follows:
u = a + b + c \\ v = ab + bc + ca \\ w = abc
Now note that u^2 - 2v = 3 and a^3 + b^3 + c^3 = u(u^2 - 3v) = u\left(\frac {9 - u^2}{2}\right).

We are to prove that w^2\left(u\cdot\frac {9 - u^2}{2} + 3w\right)\le 3.

By AM-GM, we have \sqrt [3]{abc}\le \frac {a + b + c}{3}\implies w\le \frac {u^3}{3^3}.

Hence, it suffices to prove that u^7\cdot\frac {9 - u^2}{2} + \frac {u^9}{3^2}\le 3^7.

However, by QM-AM we have \sqrt {\frac {a^2 + b^2 + c^2}{3}}\ge \frac {a + b + c}{3}\implies u\le 3 which proves the above inequality. \blacksquare

I'm very sorry but I have to leave right now.

Could someone please post a new problem?

Thank you. Smile


Fantasylover , how does u\le 3 imply

u^7\cdot\frac {9 - u^2}{2} + \frac {u^9}{3^2}\le 3^7
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