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Inequality with variables between 1/2 and 1
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Valentin Vornicu
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#1
Inequality with variables between 1/2 and 1
Romanian NMO 2006, Grade 8, Problem 4
Let a,b,c \in \left[ \frac 12, 1 \right]. Prove that 2 \leq \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b} \leq 3 .
selected by Mircea Lascu
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Last edited by Valentin Vornicu on Sat Apr 22, 2006 1:42 am; edited 1 time in total 
PostPosted: Tue Apr 18, 2006 12:28 am  Back to top 
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zanttrang
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#2
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Let f(a,b,c) = \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b}

Now fix b,c and note that f(a) is convex, so it will be maximized at an endpoint of the given interval. The same is true for f(b) and f(c). So, f(a,b,c) will be maximized for some (a,b,c) \in \{1/2, 1\} (sorry for the sloppy notation there, but I think you know what I mean). Testing these points (there are only four cases, since the expression is symmetric), we find that f(a,b,c) \leq 3.

Now take the derivative of the function with respect to one variable:

f'(a) = \frac{1}{1+c} - \frac{b + c}{(1 + a)^2} + \frac{1}{1 + b} \geq 1/9 > 0

since the expression is obviously minimized for (a,b,c) = (1/2, 1, 1). Therefore, the function is strictly increasing on the given interval, so the minimum value is given for a = 1/2. Identical results hold for f(b) and f(c), so the minimum value of f(a,b,c) is given by (a,b,c) = (1/2, 1/2, 1/2) \Rightarrow f(a,b,c) \geq 2.

The second part actually proves the result found in the first part as well, but since this result uses calculus, it's not preferable.

PostPosted: Tue Apr 18, 2006 7:23 pm  Back to top 
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Marius Damian
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#3
Solution
M.D.
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We have:
2 \leq \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b} \leq 3 \Leftrightarrow 2\leq (\frac{a}{1+c}+\frac{c}{1+a})+(\frac{b}{1+a}+\frac{a}{1+b})+(\frac{c}{1+b}+\frac{b}{1+c})\leq3.
It follows:
\frac{a}{1+c}+\frac{c}{1+a}\geq \frac{2}{3} \Leftrightarrow 2a^{2}+2c^2+a+c+(a-c)^2\geq2 \Leftrightarrow 2(a^2-\frac{1}{4})+2(a^2-\frac{1}{4})+(a-\frac{1}{2})+(a-\frac{1}{2})+(a-c)^2\geq 0 and analogues.
Simultaneous:
\frac{a}{1+c}+\frac{c}{1+a}\leq 1 \Leftrightarrow a^{2}+c^2\leq 1+ac \Leftrightarrow (1-a)(1-c)+a(1-a)+c(1-c)\geq 0 and analogues.
The conclusion is obvious.

PostPosted: Tue Apr 18, 2006 11:49 pm  Back to top 
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Virgil Nicula
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#4
This nice inequality is placed very right at the grade 8 !
Marius Damian - O.K. (a nice proof), but Zanttrang ... no (a great proof, but it isn't for the grade 8) !
Here is a short (similar approx. 75% with Marius Damian) proof.
Lemma. \{x,y\}\subset \left[\frac 12 ,1\right]\Longrightarrow
1.\blacktriangleright 2(1+xy)\le 3(x^2+y^2)+(x+y)\ .
2.\blacktriangleright x^2+y^2\le 1+xy\ .
Indeed, if we will add the following evident inequalities 2xy\le x^2+y^2 and 2\le2(x^2+y^2)+(x+y) then we obtain the first inequality and the second inequality is equivalently with the evident relation (1-x)(1-y)+x(1-x)+y(1-y)\ge 0\ .

The proof of the proposed problem.
S\equiv \sum \frac{a+b}{1+c}=\sum \left(\frac{a}{1+b}+\frac{b}{1+a}\right)\in [2,3] because :
\frac 23\le \frac{a}{1+b}+\frac{b}{1+a}\Longleftrightarrow 2[(1+ab)+(a+b)]\le 3[(a^2+b^2)+(a+b)]\Longleftrightarrow (1).
\frac{a}{1+b}+\frac{b}{1+a}\le 1\Longleftrightarrow a(1+a)+b(1+b)\le (1+a)(1+b)\Longleftrightarrow (2).

PostPosted: Wed Apr 19, 2006 1:51 am  Back to top 
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delta
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#5
I think that when posting problems from national Olympiads it shoud be a great idea to tell what is the level of the participants. Not just stating grade 8 (who knows what is grade 8 in Romania exept romanians Mr. Green ) but for example (no calculus)
This is very important exspecialy in geometry problems (for examle no trig, or no vectors etc.)

PostPosted: Wed Apr 19, 2006 8:25 am  Back to top 
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kunny
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#6
What's age for grade 8 in Romania?

PostPosted: Wed Apr 19, 2006 8:34 am  Back to top 
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Virgil Nicula
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#7
The seventh class (approx. 14years age).

PostPosted: Wed Apr 19, 2006 9:32 am  Back to top 
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wubingjie
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#8
Re: Inequality with variables between 1/2 and 1
Romanian NMO 2006, Grade 8, Problem 4
Valentin Vornicu wrote:
Let a,b,c \in \left[ \frac{1}{2}, 1 \right]. Prove that
2 \leq \frac{ a+b}{1+c}+\frac{ b+c}{1+a}+\frac{ c+a}{1+b}\leq 3 .
selected by Mircea Lascu

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PostPosted: Wed Jan 03, 2007 6:35 am  Back to top 
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mychrom
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#9
A simple ineq, even for 8th grade (i'm 8th grade in Romania).
1\leq a+b so 1+c\leq a+b+c and the first ineq is solve.
for the second one we will must show that a/1+c+c/1+a\leq 1 ( like Mr. Damian). this is equvalent to (a-c)^{2}\leq 1+ac.

PostPosted: Thu Jan 04, 2007 4:04 am  Back to top 
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mychrom
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#10
mychrom wrote:
A simple ineq, even for 8th grade (i'm 8th grade in Romania).
1\leq a+b so 1+c\leq a+b+c and the first ineq is solve.
for the second one we will must show that a/1+c+c/1+a\leq 1 ( like Mr. Damian). this is equvalent to (a-c)^{2}\leq 1+ac.


sorry there was only a^{2}+c^{2}\leq 1+ac, which is trivial. Blush

PostPosted: Thu Jan 04, 2007 4:15 am  Back to top 
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Potla
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#11
Re: Inequality with variables between 1/2 and 1
Romanian NMO 2006, Grade 8, Problem 4
Valentin Vornicu wrote:
Let a,b,c \in \left[ \frac 12, 1 \right]. Prove that
2 \leq \frac { a + b}{1 + c} + \frac { b + c}{1 + a} + \frac { c + a}{1 + b} \leq 3 .
selected by Mircea Lascu

Here is my solution:
\sum \frac {a + b}{1 + c} = \sum \frac {(a + b)^2}{a + b + ac + bc} \geq \frac {4(a + b + c)^2}{2(a + b + c + ab + bc + ca)}

= 2 \cdot \frac {(a + b + c)^2}{a + b + c + ab + bc + ca}
So we only have to prove that
(a + b + c)^2\geq a + b + c + ab + bc + ca \iff a^2 + b^2 + c^2 + ab + bc + ca\geq a + b + c

\iff a(a + b - 1) + b(b + c - 1) + c(c + a - 1)\geq 0
Which is true since 1\leq a + b;b + c;c + a.

(EDIT: My solution to the right part was not correct so I omitted it Sad Sad )
PS/Edit
Sorry for reviving this old thread, but I came to know from "Agr_94_Math" that this was Karnataka Mathematical Olympiad 2006; so I found out this solution. Smile

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PostPosted: Sun Nov 08, 2009 1:13 am  Back to top 
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