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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
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0 replies
jlacosta
Nov 1, 2024
0 replies
Symmetric geo with tangent circles
AlephG_64   1
N 2 minutes ago by sami1618
Source: 2nd AGO P4
Let $ABC$ be a scalene triangle. The perpendicular bisector of $BC$ intersects lines $AB$ and $AC$ at $A_b$ and $A_c$ respectively. Let $O_a$ denote the circumcenter of triangle $AA_bA_c$. Define $O_b, O_c$ similarly.
Prove that the circumcircle of triangle $O_aO_bO_c$ is tangent to the circumcircle of triangle $ABC$.

Proposed by Atavic
1 reply
AlephG_64
2 hours ago
sami1618
2 minutes ago
a+b+c+abc=4 with two equality cases
KhuongTrang   48
N 15 minutes ago by KhuongTrang
Source: own
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c+abc=4.$ Prove that
$$\color{blue}{\sqrt{14a+b+c} +\sqrt{14b+a+c} +\sqrt{14c+b+a}\le 2+2\sqrt{5}\cdot\sqrt{a+b+c+2}. }$$Equality holds iff $a=b=c=1$ or $a=b=2,c=0$ and any permutations.
48 replies
KhuongTrang
Mar 11, 2024
KhuongTrang
15 minutes ago
a^2+ b^2+ c^2
sqing   3
N 35 minutes ago by sqing
Source: Own
Let $ a,b, c$ be reals such that $a +b =2 , b^2+2c^2= 9 $ and $ 3c^2+4a^2 =48.$ Prove that$$a^2+ b^2+ c^2=14$$Let $ a,b ,c$ be reals such that $a +b =1 , b^2+2c^2= 9 $ and $ 3c^2+4a^2 =48.$ Prove that$$a^2+ b^2+ c^2=\frac{7(101-6\sqrt{41})}{25}$$
3 replies
sqing
Yesterday at 1:48 AM
sqing
35 minutes ago
Needed help on inequalities
fAaAtDoOoG   10
N an hour ago by arqady
Hello guys, this is my first post. I've encountered an inequality and struggled to solve it. If anyone can solve it, that would be awesome. Any help would be greatly appreciated!!!

$$\frac{b+c}{\sqrt{a^{2}+bc}} + \frac{c+a}{\sqrt{b^{2}+ac}} + \frac{a+b}{\sqrt{c^{2}+ab}} > 4, a,b,c \in \mathbb{R}^{+}$$
10 replies
fAaAtDoOoG
Yesterday at 1:28 AM
arqady
an hour ago
No more topics!
Iran TST 2009-Day1-P3
khashi70   55
N Sunday at 10:07 PM by anduran
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that :

$ \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4}$
55 replies
khashi70
May 9, 2009
anduran
Sunday at 10:07 PM
Iran TST 2009-Day1-P3
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khashi70
239 posts
#1 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that :

$ \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4}$
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Vasc
2861 posts
#16 • 6 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 2 other users
khashi70 wrote:
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that :

$ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$
The best inequality of this type is the following:

If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then

$ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)}\le \frac 1{6}$,

with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation.
This post has been edited 1 time. Last edited by v_Enhance, Sep 7, 2015, 12:34 PM
Reason: c^2+d^2 should be c^2+a^2
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can_hang2007
2948 posts
#21 • 8 Y
Y by doxuanlong15052000, HappyMathEducation, ImSh95, TheHawk, Adventure10, Mango247, and 2 other users
khashi70 wrote:
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that :

$ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$
Write the inequality as
$ \sum \frac{a^2+b^2}{a^2+b^2+2} \ge \frac{3}{2},$
or
$ \sum \frac{(a+b)^2}{(a+b)^2+\frac{2(a+b)^2}{a^2+b^2}} \ge \frac{3}{2}.$
Applying the Cauchy Schwarz Inequality, we have
$ LHS \ge \frac{4(a+b+c)^2}{\sum (a+b)^2+2\sum \frac{(a+b)^2}{a^2+b^2}} =\frac{36}{\sum (a+b)^2+2\sum \frac{(a+b)^2}{a^2+b^2}}.$
It suffices to prove that
$ \sum (a+b)^2+ 2\sum \frac{(a+b)^2}{a^2+b^2} \le 24.$
Because $ 12-\sum (a+b)^2 =\frac{4}{3} (a+b+c)^2 -\sum (a+b)^2 =-\frac{1}{3} \sum (a-b)^2,$ and $ 12-2\sum \frac{(a+b)^2}{a^2+b^2} =2\sum \frac{(a-b)^2}{a^2+b^2},$ this inequality is equivalent to
$ \sum (a-b)^2 \left( \frac{6}{a^2+b^2} -1\right) \ge 0.$
Under the assumption that $ a \ge b \ge c,$ we see that this inequality is obviously true if $ a^2+b^2 \le 6.$ Let us consider now the case $ a^2+b^2 \ge 6,$ in this case we have
$ \frac{1}{a^2+b^2+2} \le \frac{1}{8},$
and
$ \frac{1}{a^2+c^2+2}+\frac{1}{b^2+c^2+2} \le \frac{1}{a^2+2}+\frac{1}{b^2+2} \le \frac{1}{8-b^2}+\frac{1}{b^2+2} \le \frac{1}{8}+\frac{1}{2},$
(because $ 0 \le b^2 \le \frac{9}{4}$)
Hence
$ \sum \frac{1}{a^2+b^2+2} \le \frac{1}{8}+\frac{1}{8}+\frac{1}{2} =\frac{3}{4}.$
Our proof is completed. :)
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hophinhan
103 posts
#23 • 5 Y
Y by sewie, ImSh95, TheHawk, Adventure10, and 1 other user
Vasc wrote:
khashi70 wrote:
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that :

$ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$
The best inequality of this type is the following:

If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then

$ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$,

with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation.

Primitive mixing variables also can solve this one.

Solution:


WLOG $ a = max\{a,b,c\}$
\[ f(a,b,c) = \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + a^2)}
\]

\[ f(a,\frac {b + c}{2},\frac {b + c}{2}) = \frac 2{8 + 5(a^2 + \frac {(b + c)^2}{4})} + \frac 1{8 + 5\frac {(b + c)^2}{2}}
\]

\[{ \Rightarrow \ \ f(a,\frac {b + c}{2},\frac {b + c}{2}}) - f(a,b,c) =
\]

\[ (b - c)^2. \left[ \frac {\frac {5}{4}(16 + 10a^2 - 5b^2 - 5c^2 - 20bc)}{[8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)]} + \frac {\frac {5}{2}}{[8 + \frac {5(b + c)^2}{2}][8 + 5(b^2 + c^2]}\right]
\]
Note that :
\[ [8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)] \ge [8 + \frac {5(b + c)^2}{2}][8 + 5(b^2 + c^2)]^2
\]

\[{ \Rightarrow \ \ f(a,\frac {b + c}{2},\frac {b + c}{2}} - f(a,b,c)\ge
\]

\[ \ge (b - c)^2. \left[ \frac {\frac {5}{4}(16 + 10a^2 - 5b^2 - 5c^2 - 20bc) + \frac {5}{2}(8 + 5(b^2 + c^2)}{[8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)]}  \right] \ge
\]

\[ \ge (b - c)^2. \left[ \frac {\frac {5}{4}(32 + 10a^2 - 5b^2 - 5c^2 - 20bc) + \frac {25(b^2 + c^2)}{2}}{[8 + 5(a^2 + b^2)][8 + 5(a^2 + \frac {(b + c)^2}{2}][8 + 5(a^2 + c^2)]} \right] \ge 0
\]
Therefore, we only need to prove the inequality in the case when b=c . Now it's a very simple. :ninja:
This post has been edited 1 time. Last edited by hophinhan, May 12, 2009, 9:30 AM
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zaizai-hoang
584 posts
#27 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
Mixing Variables method is not enought for Vasc's problem... I think that. Because it only works for the case $ 0\le bc\le \frac{16}{20}$.
But it can show you why we have the second equality. In fact we have:
\[ f(a,b,c)=RHS-LHS\]
And
\[ f\left(x,\frac{3-x}{2},\frac{3-x}{2}\right)=\frac{5(x-1)^2(5x-13)^2}{6(25x^2-30x+77)(5x^2-30x+61)}\ge 0\]
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tuantam1lan
109 posts
#30 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
khashi70 wrote:
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that :

$ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ .find constant k is best such that:
$ \frac {1}{1 + k(a^{2} + b^{2})} + \frac {1}{1 + k(b^{2} + c^{2})} + \frac {1}{1 + k(c^{2} + a^{2})} \leq \frac {3}{1 + 2k}$
i think $ k<\frac{2}{3}$
This post has been edited 1 time. Last edited by tuantam1lan, May 13, 2009, 11:35 AM
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Gibbenergy
640 posts
#31 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
Vasc wrote:
khashi70 wrote:
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that :

$ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$
The best inequality of this type is the following:

If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then

$ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$,

with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation.

Since $ a+b+c = 3= const$. Fix $ a^2+b^2+c^2 = k = constant$
Consider function $ f=\frac{1}{8+5(k-a^2)}$ . $ f'''(a) = \frac{600a(5a^2+8+5k)}{(-8-5k+5a^2)^4} >0$ so apply n-1 EV, we only consider $ f(a,a,b)$. Now the problem is much simple and can be solved.
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alex2008
871 posts
#32 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
This Iran 2009 is an old result of Vasc

http://www.mathlinks.ro/Forum/viewtopic.php?t=77182
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quykhtn-qa1
1347 posts
#34 • 9 Y
Y by hal9v4ik, dm_edogawasonichi, Hermitianism, ImSh95, TheHawk, Schur-Schwartz, Adventure10, xyr, and 1 other user
khashi70 wrote:
Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a + b + c = 3$ . Prove that :

$ \frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}} \leq \frac {3}{4}$
Here is my solution:
$ Lemma$ :With $ x,y,z>0$,we have:
$ (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2 \ge 2\sqrt{3(xy+yz+zx)}+4(x+y+z)$
$ Proof$ .
Put $ m=\sqrt{b+c}+\sqrt{c+a}-\sqrt{a+b};n=\sqrt{c+a}+\sqrt{a+b}-\sqrt{b+c}$
$ p=\sqrt{a+b}+\sqrt{b+c}-\sqrt{c+a}$,then $ m,n,p>0$
This inequality become: $ mn+np+pm \ge \sqrt{3mnp(m+n+p)}$
Which is obvious true. Equality holds if and only if $ x=y=z$
-Now,com back the problem
It is equivalent to:
$ \frac{a^2+b^2}{2+a^2+b^2}+\frac{b^2+c^2}{2+b^2+c^2}+\frac{c^2+a^2}{2+c^2+a^2} \ge \frac{3}{2}$
By Cauchy-Schwarz inequality and Lemma,we have;
$ LHS \ge \frac{(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2}{6+2(a^2+b^2+c^2)}$
$ \ge \frac{2(a^2+b^2+c^2)+ab+bc+ca}{3+a^2+b^2+c^2}$ (because:$ \sqrt{3(a^2b^2+b^2c^2+c^2a^2} \ge ab+bc+ca$)
$ = \frac{\frac{3}{2}(3+a^2+b^2+c^2)}{3+a^2+b^2+c^2}=RHS$ (because $ a+b+c= 3$)
Then,we have Q.E.d
Equality holds if and only if $ a=b=c=1$
:)
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nguoivn
1453 posts
#35 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
Your proof can be write into:

$ LHS \geq\  \frac{(\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}})^{2}}{6+2(a^{2}+b^{2}+c^{2})}$
$ = \frac{2\sum\ a^2+2\sum\ \sqrt{(a^2+b^2)(a^2+c^2)}}{6+2(a^{2}+b^{2}+c^{2})}$ $ \geq\ \frac{2\sum\ a^2+2(\sum\ a^2+\sum\ ab)}{6+2(a^2+b^2+c^2)} = \frac{3}{2}$
@quykhtn-qa1: Click to reveal hidden text
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vo thanh van
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#36 • 6 Y
Y by thangpbc, ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
nice solution,quykhtn-a1
Here is my solution
Let $ f(a,b,c)=\frac {1}{2 + a^{2} + b^{2}} + \frac {1}{2 + b^{2} + c^{2}} + \frac {1}{2 + c^{2} + a^{2}}$
=> $ f(a,t,t)=\frac{2}{2+a^2+t^2}+\frac{1}{2+2t^2}$ with $ t=\frac{b+c}{2}$
We have $ f(a,t,t)-f(a,b,c)$
$ =(b^2+c^2-\frac{(b+c)^2}{2})(\frac{1}{(b^2+c^2+2)(2+\frac{(b+c)^2}{2})}$ $ -\frac{1}{(4+2a^2+b^2+c^2)(4+2a^2+\frac{(b+c)^2}{2})})\ge 0$
It's true because $ 2(b^2+c^2)\ge (b+c)^2$ and $ 4+2a^2+b^2+c^2\ge b^2+c^2+2,4+2a^2+\frac{(b+c)^2}{2}\ge 2+\frac{(b+c)^2}{2}$
And then,we prove that $ f(a,t,t)\le\frac{3}{4} \Leftrightarrow (a-1)^2(15a^2-78a+111)\ge 0$ (which is obvious true)(Q.E.D)
Equality holds if and only if $ a=b=c=1$
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arqady
29867 posts
#38 • 8 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 4 other users
Vasc wrote:
If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then

$ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$,

with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation.
This inequality is true for all reals $ a,$ $ b$ and $ c$ such that $ a + b + c = 3.$
Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2$ and $ abc = w^3.$
Since, $ \prod_{cyc}(a^2 + b^2) = - w^6 + A(u,v^2)w^3 + B(u,v^2)$
we see that the original inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is convex function.
Hence, $ f$ gets a maximal value, when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal.
Thus, for the poof enough to check only one case:
$ b = c = \frac {3 - a}{2},$ which gives $ (a - 1)^2(5a - 13)^2\geq0.$ Done! :)
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Karamata
30 posts
#39 • 5 Y
Y by ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
arqady wrote:
Vasc wrote:
If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then

$ \frac 1{8 + 5(a^2 + b^2)} + \frac 1{8 + 5(b^2 + c^2)} + \frac 1{8 + 5(c^2 + d^2)}\le \frac 1{6}$,

with equality for $ a = b = c$, and again for $ \frac a{13} = b = c$ or any permutation.
This inequality is true for all reals $ a,$ $ b$ and $ c$ such that $ a + b + c = 3.$
Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2$ and $ abc = w^3.$
Since, $ \prod_{cyc}(a^2 + b^2) = - w^6 + A(u,v^2)w^3 + B(u,v^2)$
we see that the original inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is convex function.
Hence, $ f$ gets a maximal value, when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal.
Thus, for the poof enough to check only one case:
$ b = c = \frac {3 - a}{2},$ which gives $ (a - 1)^2(5a - 13)^2\geq0.$ Done! :)

Maybe you mean $ f(u,v^2,w^3)\leq 0$ and $ f$ is convex?
This post has been edited 1 time. Last edited by Karamata, Mar 3, 2010, 10:43 PM
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arqady
29867 posts
#40 • 4 Y
Y by ImSh95, TheHawk, Adventure10, and 1 other user
Karamata wrote:
Maybe you mean $ f(u,v^2,w^3)\leq 0$ and $ f$ is convex? Otherwise, it doesn't follow that $ f$ will attain its maximum only if $ w^3$ is extremal :maybe:
$ f$ is convex function of $ w^3.$ :wink:
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Vasc
2861 posts
#41 • 6 Y
Y by hctb00, ImSh95, TheHawk, Adventure10, Mango247, and 1 other user
This is also my proof, Arqady. :lol:

We can generalize this inequality to $ n$ variables as follows:

Let $ a_1,a_2,...a_n$ be real numbers such that $ a_1 + a_2 + ... + a_n = n$. If $ r\ge \frac {n^2 - 1}{n^2 - n - 1}$, then

$ \sum \frac 1{r + a_2^2 + ... + a_n^2} \le \frac n{r + n - 1}$,

with equality for $ a_1 = ... = a_n = 1$, and for $ a_1 = ... = a_{n - 1} = \frac 1{n^2 - n - 1}$ and $ a_n = n - \frac {n - 1}{n^2 - n - 1}$.
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