Singapore TST 2004

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Singapore TST 2004

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#1 h May 10, 2004, 4:26 PM • 2 Y

Y by Adventure10, Mango247

Let $0 < a, b, c < 1$ with $ab + bc + ca = 1$ . Prove that
$\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2} \geq \frac {3 \sqrt{3}}{2}.$

Determine when equality holds.

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Valentin Vornicu

7301 posts

Valentin Vornicu

#2 h May 10, 2004, 10:09 PM • 2 Y

Y by Adventure10, Mango247

as it turns out the function $\displaystyle f(x) = \frac x{1-x^2}$ is convex on $(0,1)$ because $f''(x) = \frac { 4a(1-a^2)}{(1-a^2)^4} >0,$ so we can apply Jensen to obtain that

$\sum \frac a{1-a^2} \geq 3 \frac { \frac s3}{ 1 - \frac {s^2}9 } \geq \frac {3\sqrt 3}2 \Leftrightarrow$
$\frac {s^2}9 + \frac 2{3\sqrt 3} s \geq 1 \quad (1)$ where $s=a+b+c$ . But $(a+b+c)^2 \geq 3 (ab+bc+ca)=3$ , so $s\geq 3$ , so obviously (1) takes place. The equality can only follow when $a=b=c= \displaystyle \frac 1{\sqrt 3}$ .

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Myth

#3 h May 11, 2004, 5:18 AM • 2 Y

Y by Adventure10, Mango247

Another approach is following: $\frac{x}{1-x^2}\geq\frac{3\sqrt{3}}{2}\cdot x^2$ .

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#4 h May 11, 2004, 3:06 PM • 2 Y

Y by Adventure10, Mango247

When did the two of you become beginners?

And yet another simple approach is to use the trigonometric substitution $a = \tan A, b = \tan B, c = \tan C, \mbox{ where } A + B + C = \pi/2$ .

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Myth

#5 h May 11, 2004, 4:06 PM • 2 Y

Y by Adventure10, Mango247

I am not guilty, Valentin did it first.

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Valentin Vornicu

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Valentin Vornicu

#6 h May 11, 2004, 4:39 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Myth wrote:

I am not guilty, Valentin did it first.

sorry, never gonna happen again

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Palosh

#7 h May 11, 2004, 6:28 PM • 2 Y

Y by Adventure10, Mango247

Could you bring the Singapore Selection Tests 2004?
thank you very much...

Moldova rulezzzz

_________________________
"Our days are never coming back.... (System of a down -> Highway Song)"

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#8 h May 12, 2004, 3:20 PM • 2 Y

Y by Adventure10, Mango247

I posted all 6 questions from the Singapore TST 2004, so they're lying in various threads around the forum. You can just do a search to find them.

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Valentin Vornicu

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Valentin Vornicu

#9 h May 12, 2004, 3:42 PM • 2 Y

Y by Adventure10, Mango247

Maybe you can gather up all the links like I did with the Romanian TST 2004 and open up a thread on the National Olympiads forum

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#10 h May 13, 2004, 5:38 PM • 2 Y

Y by Adventure10, Mango247

By the way, Singapore TST is it just for beginners ??!!

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#11 h May 15, 2004, 5:46 PM • 2 Y

Y by Adventure10, Mango247

Obviously it's not just for beginners, it just so happens that the 1st 2 questions were rather easy (as far as I see, so that everybody's marks would look a bit better). There were harder questions such as this and that.

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2280 posts

#12 h Oct 2, 2014, 9:06 AM • 2 Y

Y by Adventure10, Mango247

It's a very old thread

But I'm going to post a different (but easy to obtain) solution

$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\geq\frac{3\sqrt{3}}{2}.$ $\iff \sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge 0$

Let $a\ge b\ge c$
$\implies 2a+\sqrt 3 a^2 - \sqrt 3 \ge 2b+\sqrt 3 b^2 - \sqrt 3 \ge 2c+\sqrt 3 c^2 - \sqrt 3$
And $1-a^2 \le 1-b^2 \le 1-c^2$

Thus by Tchebychef's inequality
$3\sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge (\sum_{cyclic}2a+\sqrt 3 a^2 - \sqrt 3)(\sum_{cyclic} \frac{1}{1-a^2})$

Since $ab+bc+ca=1 \implies a+b+c \ge \sqrt 3$ and $a^2+b^2+c^2 \ge 1$
Combining the above two results
$(\sum_{cyclic}2a+\sqrt 3 a^2 - \sqrt 3)(\sum_{cyclic} \frac{1}{1-a^2}) \ge 0$
$\implies \sum_{cyclic}\frac{2a+\sqrt 3 a^2 - \sqrt 3 }{1-a^2} \ge 0$

QED

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#13 h Oct 22, 2016, 3:07 PM • 1 Y

Y by Adventure10

Easy. Just a Cauchy-Schwartz and it is equivalent to a+b+c>=sqrt{3}, which is trivial.

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#14 h Oct 22, 2016, 3:43 PM • 1 Y

Y by Adventure10

//cdn.artofproblemsolving.com/images/7/2/c/72c26c94356bcd0379aaedd07a4be2c980b4c8b2.jpg

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xzlbq

#15 h Oct 22, 2016, 11:06 PM • 3 Y

Y by luofangxiang, Adventure10, Mango247

luofangxiang wrote:

http://s12.sinaimg.cn/middle/006ptkjAzy75Pleo5GX3b&690

easy.
see this:

${\frac {\sqrt {{x}^{2} \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }}{{y}^{ 2}+{z}^{2}}}\geq 3/2\,{\frac {\sqrt {3}{x}^{2}}{{x}^{2}+{y}^{2}+{z}^{2}}}$
equality hold when $2x^2=y^2+z^2$

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#16 h Nov 28, 2016, 1:50 PM • 1 Y

Y by Adventure10

Similar, generalisated: sum a/(b^2+c^2)>=9/2sqrt(a^2+b^2+c^2).

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F10tothepowerof34

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F10tothepowerof34

#17 h Jun 26, 2023, 2:22 PM

Y by

Let $f(x)=\frac{x}{1-x^2}\text{ furthermore notice that }f''(x)=\frac{4x(1-x^2)}{(1-x^2)^4}>0\text{ for } x\in(0,1)$ thus the function is convex on this interval.
Furthermore $\sum_{cyc}\frac{a}{1-a^2}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f\left(\frac{\sum_{cyc}a}{3}\right)=\frac{\sum_{cyc}a}{1-\frac{\left(\sum_{cyc}a\right)^2}{9}}$
Therefore the inequality boils down to $\frac{\sum_{cyc}a}{1-\frac{\left(\sum_{cyc}a\right)^2}{9}}\ge\frac{3\sqrt{3}}{2}\Longrightarrow2\sum_{cyc}a\ge3\sqrt{3}-\frac{\sqrt{3}\left(\sum_{cyc}a\right)^2}{3}\Longleftrightarrow\left(\sum_{cyc}a\right)\left(6+\sqrt{3}\sum_{cyc}a\right)\ge9\sqrt{3}$ , which is clearly true as $\left(\sum_{cyc}a\right)^2\ge3\sum_{cyc}ab\Longrightarrow \sum_{cyc}a\ge\sqrt{3}$ $\blacksquare$ .

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