Inequality with square roots

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Inequality with square roots

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Source: ELMO Shortlist 2012, A1

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4302 posts

#1 h Jul 2, 2012, 3:07 AM • 2 Y

Y by Adventure10, Mango247

Let $x_1,x_2,x_3,y_1,y_2,y_3$ be nonzero real numbers satisfying $x_1+x_2+x_3=0, y_1+y_2+y_3=0$ . Prove that
$\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}+\frac{x_2x_3+y_2y_3}{\sqrt{(x_2^2+y_2^2)(x_3^2+y_3^2)}}+\frac{x_3x_1+y_3y_1}{\sqrt{(x_3^2+y_3^2)(x_1^2+y_1^2)}} \ge -\frac32.$

Ray Li, Max Schindler.

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41002 posts

sqing

#2 h Jul 3, 2012, 5:23 AM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

Let $a,b,c$ be nonnegative real numbers. Prove that
$a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.$
（ the 2009 Entirely Legitimate Junior Mathematical Olympiad ）
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&p=1563959.)
In terms of $n\ge2$ , find the largest constant $c$ such that for all nonnegative $a_1,a_2,\ldots,a_n$ satisfying $a_1+a_2+\ldots+a_n=n$ , the following inequality holds:
$\frac1{n+ca_1^2}+\frac1{n+ca_2^2}+\cdots+\frac1{n+ca_n^2}\le \frac{n}{n+c}.$
（ELMO Shortlist 2011, A4）
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487111&p=2729517#p2729517
Given positive reals $x,y,z$ such that $xy+yz+zx=1$ , show that
$\sum_{\text{cyc}}\sqrt{(xy+kx+ky)(xz+kx+kz)}\ge k^2,$ where $k=2+\sqrt{3}$ .
（ELMO Shortlist 2011, A5）
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=487112&p=2729518#p2729518
Let $a,b,c$ be three positive real numbers such that $a \le b \le c$ and $a+b+c=1$ . Prove that
$\frac{a+c}{\sqrt{a^2+c^2}}+\frac{b+c}{\sqrt{b^2+c^2}}+\frac{a+b}{\sqrt{a^2+b^2}} \le \frac{3\sqrt{6}(b+c)^2}{\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}.$
（ELMO Shortlist 2012, A2）
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728453&sid=6827ef4989b84c04da643548d9fa2134#p2728453
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=239&year=2012
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=239&year=2011

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meowme

#3 h Jul 5, 2012, 1:48 PM • 3 Y

Y by Siddharth03, Adventure10, Mango247

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john10

#4 h Jan 30, 2016, 11:04 PM • 1 Y

Y by Adventure10

Is there another solution for this nice problem ?

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#5 h Feb 24, 2017, 3:30 PM • 3 Y

Y by Siddharth03, Adventure10, Mango247

My solution.
$x_1+x_2+x_3= 0 \implies x_1x_2= \frac{x_3^2-x_1^2-x_2^2}{2}$
$y_1+y_2+y_3= 0 \implies y_1y_2= \frac{y_3^2-y_1^2-y_2^2}{2}$
$\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}=-\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)-(x_3^2+y_3^2)}{2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}$
Set $a=\sqrt{x_1^2+y_1^2},b=\sqrt{x_2^2+y_2^2},c=\sqrt{x_3^2+y_3^2}$
We need to prove
$\sum{\frac{a^2}{bc}}-\sum{\frac{a}{b}}-\sum{\frac{b}{a}}+3 \ge 0$
<-> $\frac{\sum{a^3}+3abc-\sum{a^2b+b^2a}}{abc} \ge 0$
Obviously true by schur

This post has been edited 1 time. Last edited by MonsterS, Mar 9, 2017, 3:20 PM

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#6 h Feb 24, 2017, 10:20 PM • 4 Y

Y by Pure_IQ, kapilpavase, Adventure10, Mango247

math154 wrote:

Let $x_1,x_2,x_3,y_1,y_2,y_3$ be nonzero real numbers satisfying $x_1+x_2+x_3=0, y_1+y_2+y_3=0$ . Prove that
$\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}+\frac{x_2x_3+y_2y_3}{\sqrt{(x_2^2+y_2^2)(x_3^2+y_3^2)}}+\frac{x_3x_1+y_3y_1}{\sqrt{(x_3^2+y_3^2)(x_1^2+y_1^2)}} \ge -\frac32.$
Ray Li, Max Schindler.

Beautiful problem!!
Let us consider three vectors $a = (x_{1} ,y_{1}) ,b = (x_{2} ,y_{2}),a = (x_{1} ,y_{1})$ .
See that $a+b+c =(0,0)$ . So the three vectors form a triangle. Let $A,B,C$ be the angles between these vectors.
We need to show that
$cosA + cosB + cosC \ge \frac{-3}{2}$ , where $A+B+C = 2\pi$

Let $cos\frac{A}{2} = x , cos\frac{B}{2}= y , cos\frac{C}{2} = z =>x^2 + y^2 + z^2 + 2xyz = 1$

Claim : $x^2 + y^2 +z^2 \ge\frac{3}{4}$
Proof :
If $x^2 +y^2 +z^2 < \frac{3}{4} => (|xyz|)^{\frac{2}{3}} \le \frac{(x^2 +y^2 +z^2)}{3} < \frac{1}{4}=>x^2 +y^2 +z^2 +2xyz < \frac{3+1}{4} = 1$ ,contradiction.

So, $cosA + cosB + cosC + 3 = 2(cos^2\frac{A}{2} + cos^2\frac{B}{2} +cos^2\frac{C}{2}) \ge \frac{3}{2}$ Done!!

This post has been edited 2 times. Last edited by SAUDITYA, Feb 28, 2017, 8:53 AM

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30147 posts

arqady

#7 h Mar 9, 2017, 1:34 PM • 2 Y

Y by Adventure10, Mango247

MonsterS wrote:

$\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}=\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)-(x_3^2+y_3^2)}{2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}$

I think it should be $\frac{x_1x_2+y_1y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}=-\frac{(x_1^2+y_1^2)+(x_2^2+y_2^2)-(x_3^2+y_3^2)}{2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}$ .
Nice solution!

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#8 h Mar 9, 2017, 3:19 PM • 2 Y

Y by Adventure10, Mango247

Thank you.I will edit it.

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