Iran Team Selection Test 2016

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Source: TST 2,day 1,P2

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#1 h Jul 15, 2016, 7:32 PM • 3 Y

Y by Adventure10, Mango247, ehuseyinyigit

Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$ . Prove that $\sum_{cyc} \sqrt{\frac{a^2+1}{2}} \geq (3.\sum_{cyc} \sqrt{a}) -8$

This post has been edited 1 time. Last edited by MRF2017, Jul 15, 2016, 8:33 PM

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sqing

#2 h Jul 15, 2016, 10:10 PM • 7 Y

Y by Mathskidd, ali.agh, MariusStanean, Nathanisme, Adventure10, Mango247, ehuseyinyigit

MRF2017 wrote:

Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$ . Prove that $\sum_{cyc} \sqrt{\frac{a^2+1}{2}} \geq (3.\sum_{cyc} \sqrt{a}) -8$

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arqady

#3 h Jul 17, 2016, 5:55 PM • 2 Y

Y by Adventure10, Mango247

MRF2017 wrote:

Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$ . Prove that $\sum_{cyc} \sqrt{\frac{a^2+1}{2}} \geq (3.\sum_{cyc} \sqrt{a}) -8$

$\sum_{cyc}\left(\sqrt{\frac{a^2+1}{2}}-3\sqrt{a}+2\right)=\sum_{cyc}\left(\sqrt{\frac{a^2+1}{2}}-3\sqrt{a}+2-4\left(\frac{1}{a+1}-\frac{1}{2}\right)\right)$ .
Let $a=x^2$ , where $x>0$ , and $x^2+1=2ux$ .
Hence, $u\geq1$ and it remains to prove that $\sqrt{\frac{x^4+1}{2}}-3x+2-4\left(\frac{1}{x^2+1}-\frac{1}{2}\right)\geq0$ or
$(x^2+1)\sqrt{\frac{x^4+1}{2}}\geq x(3x^2-4x+3)$ or $u\sqrt{2u^2-1}\geq3u-2$ or $(u-1)^2(u^2+2u-2)\geq0$ . Done!

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sqing

#5 h Dec 14, 2016, 9:12 AM • 1 Y

Y by Adventure10

Let $a,b,c,d$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=4$ . Prove that $\sqrt[3]{\frac{a^3+b^3}{2}}+ \sqrt[3]{\frac{b^3+c^3}{2}} +\sqrt[3]{\frac{c^3+d^3}{2}}+\sqrt[3] {\frac{d^3+a^3}{2}} \leq 2(a+b+c+d)-4$

This post has been edited 1 time. Last edited by sqing, Dec 14, 2016, 9:37 AM

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#6 h Apr 22, 2017, 5:26 PM • 2 Y

Y by Adventure10, Mango247

It is a beautiful inequality. In fact the transformation is so skillful, not such easy to notice.

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#7 h Aug 8, 2018, 3:06 PM • 4 Y

Y by Lantien.C, Adventure10, Mango247, ehuseyinyigit

by changing variables $\frac{1}{a+1} \implies a=\frac{1-x}{x}$ and we define $y,z,t$ similarly and it implies $\sum x=2$ and now for the function
$f=\sqrt{\frac{2x^2-2x+1}{2x^2}}-3\sqrt{\frac{1-x}{x}}$ is convex (didn't do it by hand, used geogebra

) so the minimum is obtained when all variable are equal

This post has been edited 5 times. Last edited by H.HAFEZI2000, Aug 8, 2018, 3:42 PM

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#8 h Apr 7, 2025, 3:02 PM

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Really awful and dirty problem(at least my solution).

I like approach of the @above and I tried to prove it but it's really hard to do by hand. So I look for another approach like #3 by approximate $f(x)=\sqrt{\frac{x^2+1}{2}}-3\sqrt{x}$ by $g(x)=\frac{a}{x+1}+b$ at $x=1$ (equality case). So we can find $a=4$ and $b=-4$ by two eqautions:
$f(1)=g(1)$ and $f'(1)=g'(1)$ . We just need to prove $g(x)\leq f(x)$ . And for that just replace $x$ by $x^2$ for simplicity and just bash it. it took about a half of hour for me. Or you can use a more clever tricks like #3.

This post has been edited 2 times. Last edited by amirhsz, Apr 7, 2025, 3:06 PM

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