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3377 posts

mudok

#1 h Oct 18, 2012, 5:14 AM • 2 Y

Y by Adventure10, Mango247

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$

Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$

USAMO 2001

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3251 posts

manlio

#2 h Oct 22, 2012, 11:18 AM • 1 Y

Y by Adventure10

Very nice inequality!!

It is equivalent to $\sum a^2 +abc\sum a \geq 2\sum ab$ with $\sum a^2 +abc =4$

and we know (old mudok's inequality) that $\sum a^2 +3abc \geq 2\sum ab$

so if it were $\sum a \geq 3$ it would be proved but we know that it must be $\sum a <3$

!!

Has anyone a nice idea?

Thank you very much.

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manlio

#3 h Oct 27, 2012, 4:00 PM • 2 Y

Y by Adventure10, Mango247

Mudok's inequality is equivalent to

$\sum \sqrt{yz(y+z)}[(x+y)(y+z)(z+x) +8xyz -4x(x+y)(x+z)] \geq 0$

but how to solve it ?

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#4 h Nov 2, 2012, 10:15 AM • 1 Y

Y by Adventure10

mudok wrote:

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$

Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$

USAMO 2001

It can prove by using the inequality
$c+\dfrac{(a+b)^2}{4} \leq 2.$
Manlio said that your proof only by two lines.It is very amazing.Can you post your solution?
My solution about 15 lines.

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3251 posts

manlio

#5 h Nov 2, 2012, 2:48 PM • 2 Y

Y by Adventure10, Mango247

Dear quykhtn-qa1,

when I said it exists a two lines I was only searching it

as I misunderstood what Mudok said to me. Mudok said to me he had found a "few lines" solution and (as my English is bad

) I translated "few lines" as "two or three lines" so I spent a lot of time (night and days) in searching such a miraculous solution, but yesterday I realized that such a solution doesn't exist when Mudok said to me that his solution is about 20 lines.

Very sorry. I apologize with you.

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41334 posts

sqing

#6 h Nov 3, 2012, 5:51 AM • 1 Y

Y by Adventure10

Let $a, b, c \geq 0$ and satisfy $a^2+b^2+c^2 +abc = 4 .$ Show that $ab + bc + ca - abc \leq 2.$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&p=96705

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3377 posts

mudok

#7 h Nov 3, 2012, 6:12 AM • 2 Y

Y by Adventure10, Mango247

manlio wrote:

I realized that such a solution doesn't exist when Mudok said to me that his solution is about 20 lines.

My solution is maybe more than 20 lines if I write all details. But if I write in short form by using well-known results, it is only 3-4 lines.
For example here mine and Manlio's solutions are exactly same, but there is a difference in number of lines.

Number of lines of solutions is not important

Let's see your solutions, then I will post my own

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328 posts

#8 h Nov 3, 2012, 10:40 AM • 3 Y

Y by mudok, Adventure10, Mango247

mudok wrote:

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$

Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$

USAMO 2001

Here is my solution:
WLOG we can suppose $a=\min\{a,b,c \}$ , so that: $0 \le a \le 1$
Let $b+c=s$ and $bc=t$ then we have: $s^2-(2-a)t=4-a^2 \Rightarrow t=\frac{a^2+s^2-4}{2-a}$ ********(1)
Obviously $s \ge \sqrt{4-a^2}$ , and $t \le \frac{s^2}{4} \Leftrightarrow s \le 2\sqrt{2-a}$
From (1), we can rewrite: I.E $\Leftrightarrow as+\frac{a^2+s^2-4}{2-a}-\frac{a+s-1}{2}.a.\frac{a^2+s^2-4}{2-a}-2 \le 0$
$\Leftrightarrow -as^3+(2+a-a^2)s^2+a(8-2a-a^2)s-a^4+a^3+6a^2-8a-16 \le 0$ (2)
Let $g(s)=RHS(2)$
then $g'(s)=-3as^2+2(2+a-a^2)s+a(8-2a-a^2)$
$=-3as^2+2(a+1)(2-a)s+a(2-a)(a+4)$ ,
$\ge -3as^2+2(1+a)\frac{s^3}{4}+a(4+a)\frac{s^2}{4}$ (because $2-a \ge \frac{s^2}{4}$ )
$=\frac{s^2}{4} \Big( 2(1+a)s-(8a-a^2)\Big) \ge 0$
Because it $\Leftrightarrow s \ge \frac{8a-a^2}{2(1+a)}$ , But we can prove: $s \ge \sqrt{4-a^2} \ge \frac{8a-a^2}{2(1+a)}$ , for all $0 \le a < 1$
Then $g(s)$ is creasing function, so that we need to check: $g(2\sqrt{2-a}) \le 0$
It will be easier if we use again: $b=c=\sqrt{2-a}$
Equality holds when $a=b=c=1$ or $a=0,b=c=\sqrt{2}$

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3377 posts

mudok

#9 h Nov 9, 2012, 11:48 AM • 2 Y

Y by Adventure10, Mango247

Quykhtn-qa1, is your solution also by calculus?

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2130 posts

Sayan

#10 h Nov 9, 2012, 2:39 PM • 4 Y

Y by Mirus, Adventure10, Mango247, mudok

Easy to see that $a,b,c \le 2$ . We substitute $a=2\cos A, b=2\cos B, c=2\cos C$ . with $0 \le A,B,C \le \frac{\pi}2$ . This gives
$\sum \cos^2 A+2\prod \cos A=1 \implies A+B+C=\pi$
Let $s, r, R$ be the classical elements of triangle $ABC$ , then the inequality is equivalent to
$\frac{s^2+r^2}{R^2}-4-\left(\frac{2r}{R}+1\right)\cdot\frac{s^2-(r+2R)^2}{R^2} \le 2$
$\iff R(s^2+r^2)-(2r+R)(s^2-(r+2R)^2) \le 6R^3$
$\iff 6R^3+2rs^2-(r+2R)^2(2r+R) \ge Rr^2$
by blundon's $\boxed{s^2 \ge 2R^2+10Rr-r^2-2(R-2r)\sqrt{R(R-2r)}}$ inequality, it is enough to show
$6R^3+2r(2R^2+10Rr-r^2)-4r(R-2r)\sqrt{R(R-2r)} \ge Rr^2+(r+2R)^2(2r+R)$
$\iff 2(R-2r)(R-r)^2 \ge 4r(R-2r)\sqrt{R(R-2r)}$
$\iff (R-r)^2 \ge 2r\sqrt{(R-r)^2+r^2}$
$\iff (R-r)^4 \ge 4r^2(R-r)^2+4r^4$
$\iff ((R-r)^2-2r^2)^2 \ge 0$
which is true! Equality holds when $R=2r$ and again when $R=(1+\sqrt2)r$ i.e. when $a=b=c=1$ and $a=\sqrt2, b=\sqrt2, c=0$ and its cyclic permutations. Hence the proof is complete. The length of the proof can be shortened if we omit the calculation step.

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1347 posts

#11 h Nov 9, 2012, 3:31 PM • 5 Y

Y by legendking723, dm_edogawasonichi, Adventure10, Mango247, mudok

mudok wrote:

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$
Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$

Without loss of generality , assume that $2 \geq a \geq b \geq c \geq 0$ .
Since $\begin{aligned} 4=a^2+b^2+c^2+abc \ & \geq a^2+b^2+c^2+abc-\dfrac{(2-c)(a-b)^2}{4} \\ & =c^2+\dfrac{(a+b)^2}{2}+\dfrac{c(a+b)^2}{4}.\end{aligned}$
We obtain $\\ \dfrac{(c+2)(a+b)^2}{4} \leq 4-c^2$ , or $\\ c+\dfrac{(a+b)^2}{4} \leq 2.$
Now,rewrite the inequality as the follows
$c(a+b)+ab\left[1-\dfrac{c(a+b+c-1)}{2}\right] \leq 2.$
By the AM-GM inequality, we have
$ab \leq \dfrac{(a+b)^2}{4} ,\ a+b+c \leq \dfrac{(a+b)^2}{4}+1+c \leq 3.$
Therefore $1-\dfrac{c(a+b+c-1)}{2} \geq 1-c \geq 0 .$
It suffices to show that
$2ct+t^2\left[1-\dfrac{c(c+2t-1)}{2}\right] \leq 2,$
or
$f(c)=2ct+t^2-\dfrac{ct^2(c+2t-1)}{2}-2 \leq 0.$
With $t=\dfrac{a+b}{2},\ c+t^2 \leq 2$ .
We have $f'(c)=2t-\dfrac{t^2(2c+2t-1)}{2}=\dfrac{t(4+t-2ct-2t^2)}{2}$
Because
$4+t-2ct-2t^2 \geq 4+t-2t(2-t^2)-2t^2=2(t^2-1)(t-1)+2-t>0$
Therefore $f(c) \leq f(2-t^2)=-\dfrac{t^6-2t^5-3t^4+8t^3-8t+4}{2}=-\dfrac{(t-1)^2(t^2-2)^2}{2} \leq 0.$
Equality holds when $a=b=c=1$ or $a=b=\sqrt{2},\ c=0$ and any cyclic permutations.

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41334 posts

sqing

#12 h Nov 11, 2012, 12:11 AM • 1 Y

Y by Adventure10

If $a , b, c \ge 0$ such that $a^2+b^2+c^2+abc = 4$ prove that $a+b+c-abc \geq 2$ .
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2845215
$2+abc \le a+b+c\leq 3$ .

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3377 posts

mudok

#13 h Nov 11, 2012, 2:38 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Thank you all for your solutions. As I said before I am posting my solution:
By the condition $a+b+c\le 3$ is old result. So , $a^2+b^2+c^2+\frac{3abc}{a+b+c}\ge 4$ . Given inequality is $a^2+b^2+c^2+abc(a+b+c)\ge 2(ab+bc+ca)$ . It is enough to prove $a^2+b^2+c^2+\frac{4abc(a+b+c)}{a^2+b^2+c^2+\frac{3abc}{a+b+c}}\ge 2(ab+bc+ca)\iff \sum a^3(a-b)(a-c)\ge 0$ which is Schur's inequality!

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1093 posts

#14 h Nov 25, 2012, 9:01 AM • 1 Y

Y by Adventure10

Inequality is equivalent to:
$a^2+b^2+c^2+abc(a+b+c) \geq\ 2(ab+bc+ca)$ (1)
Because $a^2+b^2+c^2+abc=4$ is equivalent to $2a+bc=\sqrt{(4-b^2)(4-c^2)}$ result by applying GM-AM
relationship $2a+bc\le\frac{8-b^2-c^2}{2}$ which is equivalent to :
$ 4a+(b+c)^2 \le\8 $ (2)
Because $4bc \le (b+c)^2$ of (2) results:
$a+bc \le\ 2$ (3)
and two other similar relationships $b+ca \le 2$ and $c+ab \le 2$
These involve the relationship
$abc \le (2-ab)(2-bc)(2-ca)$ (4)
After calculation this last relation is rewritten in the form:
$2abc(a+b+c) \geq abc+a^2b^2c^2-8+4(ab+bc+ca)$ (4')
Because equation (1) can be written as equivalent:
$2abc(a+b+c) \geq 4(ab+bc+ca)-8+ 2abc$ (1')
to prove (1) it suffices to show that:
$4(ab+bc+ca)-8+2abc \geq abc+a^2b^2c^2-8+4(ab+bc+ca)$ (5)
returns to $2abc \geq abc+a^2b^2c^2$ and finally returns to $abc\le1$
which is well-known.The demonstration ends here!

This post has been edited 6 times. Last edited by sandumarin, Feb 25, 2013, 8:07 AM

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3251 posts

manlio

#15 h Nov 25, 2012, 10:06 AM • 2 Y

Y by Adventure10, Mango247

Sandumarin, please check your solution!!

I think last step is wrong. I obtained $abc \geq 1$

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1093 posts

#16 h Nov 26, 2012, 7:46 AM • 2 Y

Y by Adventure10, Mango247

Manlio dear, you're right ,my proof is wrong! I'm trying to find a good one!
Thank you very much!

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younesmath2012maroc

621 posts

younesmath2012maroc

#17 h Feb 24, 2013, 1:26 AM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

Thank you all for your solutions. As I said before I am posting my solution:
By the condition $a+b+c\le 3$ is old result. So , $a^2+b^2+c^2+\frac{3abc}{a+b+c}\ge 4$ . Given inequality is $a^2+b^2+c^2+abc(a+b+c)\ge 2(ab+bc+ca)$ . It is enough to prove $a^2+b^2+c^2+\frac{4abc(a+b+c)}{a^2+b^2+c^2+\frac{3abc}{a+b+c}}\ge 2(ab+bc+ca)\iff \sum a^3(a-b)(a-c)\ge 0$ which is Schur's inequality!

dear ''mudok'' can you explain your solution please!!!
can you detailled please !!!

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1093 posts

#18 h Apr 11, 2013, 8:09 AM • 1 Y

Y by Adventure10

My solution:
We have:
$ab+bc+ca-abc\frac{(a+b+c-1)}{2}\leq 2\Leftrightarrow 2\sum{bc}-abc(a+b+c)+abc\leq 4\Leftrightarrow$
$\Leftrightarrow 2\sum{bc}-abc\sum{a}\leq 4-abc\Leftrightarrow 2\sum{bc}-abc\sum{a}\leq\sum{a^2}\Leftrightarrow 4\sum{bc}-abc\sum{a}\leq (\sum{a})^2\Leftrightarrow$
$\Leftrightarrow (a+b+c)^2+abc(a+b+c)\geq 4(ab+bc+ca)$
Denote $x = 2-a, y = 2-b, z = 2-c, p = x + y + z, q = xy + yz + zx, r = xyz$
Result
$\sum{a}=6-\sum{x}=6-p,\sum{bc}=\sum{(2-y)(2-z)}=12-4p+q,abc=\prod(2-x)=8-4p+2q-r$
Inequality shown returns to:
$(a+b+c)^2+abc(a+b+c)\geq 4(bc+ca+ab)\Leftrightarrow (6-p)^2+(8-4p+2q-r)(6-p)\geq 4(12-4p+q)$
Since $a^2+b^2+c^2+abc=4\Leftrightarrow\prod(2-a)=(a+b+c-2)^2\Rightarrow\prod(2-a)=r=(4-p)^2$
Inequality shown returns to:
$p^3-9p^2+36p-60+2q(4-p)\geq 0$ (1)
It is known:
$2<a+b+c\leq 3\Leftrightarrow 3\leq p<4$
$(xy+yz+zx)^2\geq 3xyz(x+y+z)\Leftrightarrow q^2\geq 3rp\Leftrightarrow q^2\geq 3p(4-p)^2$
$q^2\geq 3p(4-p)^2 ,p\geq 3\Rightarrow q^2\geq 9(4-q)^2\Leftrightarrow q\geq 3(4-p)$
For. as (1) to be true is enough to:
$p^3-9p^2+36p-60+2q(4-p)\geq p^3-9p^2+36p-60+6(4-p)^2\geq 0\Leftrightarrow p^3-9p^2+36p-60+6(16-8p+p^2)\geq 0\Leftrightarrow p^3-3p^2-12p+36\geq 0\Leftrightarrow (p-3)(p^2-12)\geq 0\Leftrightarrow p^2-12\geq 0\Leftrightarrow p^2\geq 12\Leftrightarrow p\geq 2\sqrt{3}$
In conclusion, the inequality (1) is true for $2\sqrt{3}\leq p<4$ ;
remains to prove that it is true and for $3\leq p\leq 2\sqrt{3}$
$(x+y+z)(xy+yz+zx)\geq 9xyz\Leftrightarrow pq\geq 9r\Leftrightarrow pq\geq 9(4-p)^2$
But $3\leq p\leq 2\sqrt{3}\Rightarrow 2\sqrt{3}q\geq pq\geq 9(4-p)^2\Rightarrow q\geq\frac{9}{2\sqrt{3}}(4-p)^2\Leftrightarrow q\geq\frac{3\sqrt{3}}{2}(4-p)^2$
For (1) to be true when $3\leq p\leq 2\sqrt{3}$ it is sufficient that:
$p^3-9p^2+36p-60+3\sqrt{3}(4-p)^3\geq 0$
After calculations, this last relationship turns:
${ f(p)=(3\sqrt{3}-1)p^3-(36\sqrt{3}-9)p^2-(144\sqrt{3}-36}p-192\sqrt{3}+60\leq 0$
Function f (p) is decreasing in conclusion $f(p)\leq f(3)$ ;since $f(3)=-876\sqrt{3}+222<0$
demonstration over!
_____________
Sandu Marin

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younesmath2012maroc

621 posts

younesmath2012maroc

#19 h May 10, 2013, 11:04 PM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

Thank you all for your solutions. As I said before I am posting my solution:
By the condition $a+b+c\le 3$ is old result. So , $a^2+b^2+c^2+\frac{3abc}{a+b+c}\ge 4$ . Given inequality is $a^2+b^2+c^2+abc(a+b+c)\ge 2(ab+bc+ca)$ . It is enough to prove $a^2+b^2+c^2+\frac{4abc(a+b+c)}{a^2+b^2+c^2+\frac{3abc}{a+b+c}}\ge 2(ab+bc+ca)\iff \sum a^3(a-b)(a-c)\ge 0$ which is Schur's inequality!

dear ''mudok'' can you explain your solution please!!!

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younesmath2012maroc

621 posts

younesmath2012maroc

#20 h May 14, 2013, 10:52 PM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

Thank you all for your solutions. As I said before I am posting my solution:
By the condition $a+b+c\le 3$ is old result. So , $a^2+b^2+c^2+\frac{3abc}{a+b+c}\ge 4$ . Given inequality is $a^2+b^2+c^2+abc(a+b+c)\ge 2(ab+bc+ca)$ . It is enough to prove $a^2+b^2+c^2+\frac{4abc(a+b+c)}{a^2+b^2+c^2+\frac{3abc}{a+b+c}}\ge 2(ab+bc+ca)\iff \sum a^3(a-b)(a-c)\ge 0$ which is Schur's inequality!

dear ''mudok'' can you explain your solution please!!!
i don't understand it !!!

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30161 posts

arqady

#21 h May 15, 2013, 4:11 AM • 1 Y

Y by Adventure10

younesmath2012maroc, I think the mudok's solution is true and the best here.

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3377 posts

mudok

#22 h May 15, 2013, 4:14 AM • 1 Y

Y by Adventure10

younesmath2012maroc, It is very obvious solution. I sent you private message with more details. Haven't you seen it?

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younesmath2012maroc

621 posts

younesmath2012maroc

#23 h May 15, 2013, 11:13 AM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

younesmath2012maroc, It is very obvious solution. I sent you private message with more details. Haven't you seen it?

i see it now thank's you very much

thank's !!!

This post has been edited 1 time. Last edited by younesmath2012maroc, May 15, 2013, 7:19 PM

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younesmath2012maroc

621 posts

younesmath2012maroc

#24 h May 15, 2013, 11:17 AM • 2 Y

Y by Adventure10, Mango247

i'm sorry !!!

This post has been edited 1 time. Last edited by younesmath2012maroc, May 15, 2013, 7:20 PM

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30161 posts

arqady

#25 h May 15, 2013, 11:43 AM • 2 Y

Y by Adventure10, Mango247

younesmath2012maroc wrote:

excuse me but we dont have :
$a^2+b^2+c^2+\frac{4}{a^2+b^2+c^2+\frac{3abc}{a+b+c}}\ge 2(ab+bc+ca)$

What is it?

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younesmath2012maroc

621 posts

younesmath2012maroc

#26 h May 15, 2013, 9:54 PM • 1 Y

Y by Adventure10

arqady wrote:

younesmath2012maroc wrote:

excuse me but we dont have :
$a^2+b^2+c^2+\frac{4}{a^2+b^2+c^2+\frac{3abc}{a+b+c}}\ge 2(ab+bc+ca)$

What is it?

nothing i'm sorry i have a mistake!!!

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7 posts

Cico

#27 h May 16, 2013, 11:33 AM • 2 Y

Y by Adventure10, Mango247

$a^2+b^2+c^2+abc=4$ is eqiuvalent to $(a,b,c)=(2 cos A, 2 cos B, 2 cos C)$ , where A, B, C are the angles of an acute triangle, and $A,B,C\in (0,\frac{\pi}{2})$ .
$ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2\Longleftrightarrow$
$2\cos A\cos B+2\cos B\cos C+2\cos C\cos A-2\cos A\cos B\cos C(2\cos A+2\cos B+2\cos C-1)\le 1.$
But $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1.$ So the inequality is equivalent to:
$\cos^2 A+cos^2 B+cos^2 C+(cos^2 A+cos^2 B+cos^2 C-1)(2\cos A+2\cos B+2\cos C-1)\le 1\Longleftrightarrow$
$(\cos A+\cos B+\cos C)(1-\cos^2A+\cos^2B+\cos^2C)\ge 0\Longleftrightarrow$
$(\cos A+\cos B+\cos C)\cos A\cos B\cos C\ge 0,$ which is true in any acute triangle.

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#28 h Jun 21, 2013, 9:06 AM • 2 Y

Y by Adventure10, Mango247

There's a very nice solution:
the inequality is equal to
a^2+b^2+c^2>=ab(2-c^2)+ac(2-b^2)+bc(2-a^2)
let a=2cosA, then 2-a^2=2cos(180-2A).
so the inequality is equal to
a^2+b^2+c^2>=2cos(180-2C)ab+2cos(180-2B)ac+2cos(180-2A)bc

Edit:This solution belongs to LingFu Zhang who is in the 2013 IMO China team.

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751 posts

#29 h May 31, 2015, 11:43 PM • 2 Y

Y by Adventure10, Mango247

You can prove more: you can show that the RHS is $\le 2 - \frac{(x-y)^2(y-z)^2(z-x)^2}{8}$ .

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#30 h Mar 21, 2016, 6:25 AM • 2 Y

Y by mudok, Adventure10

mudok wrote:

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$
Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$
USAMO 2001

//cdn.artofproblemsolving.com/images/8/3/8/83891320eac56d4cd2ff21fd802859f6ced1594c.jpg

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mudok

#31 h Mar 21, 2016, 7:16 AM • 1 Y

Y by Adventure10

Why is the last row true?

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#32 h Mar 21, 2016, 12:27 PM • 3 Y

Y by mudok, Adventure10, Mango247

mudok wrote:

Why is the last row true?

The first line has a hypothesis a>=b>=c

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mudok

#33 h Mar 21, 2016, 2:25 PM • 2 Y

Y by Adventure10, Mango247

Very nice proof, luofangxiang !

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mihaig

#34 h Aug 9, 2024, 11:37 PM

Y by

mudok wrote:

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$
Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$
USAMO 2001

We only have two cases:
1) $c=0$ and
2) $a=b =x,$ where $1\leq x<\sqrt2$ and $c=2-x^2.$

This post has been edited 1 time. Last edited by mihaig, Aug 9, 2024, 11:38 PM

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DipoleOfMonorak

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DipoleOfMonorak

#35 h Aug 10, 2024, 10:40 AM

Y by

mudok wrote:

$a,b,c$ are non-negative reals such that $a^2+b^2+c^2+abc=4$
Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$
USAMO 2001

it is enough to show that
$F(a,b,c)=3\Bigl(4-2(ab+bc+ca)+(a+b+c-1)abc\Bigr) + \Bigl(a^2+b^2+c^2+abc - 4\Bigr)\Bigl(a^2+b^2+c^2-ab-bc-ca +1\Bigr)\geq 0,$
but $F$ is linear by $abc$ and $F(a,b,0)\geq 0, ~F(a,a,b)\geq 0$
proof

$a,b,c$ are reals such that $a^2+b^2+c^2+abc\leq4,~c^2<4$
Prove that $ab+bc+ca-\frac{a+b+c-1}{2}\cdot abc\le 2$

This post has been edited 1 time. Last edited by DipoleOfMonorak, Aug 10, 2024, 10:46 AM

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