JBMO Shortlist 2023 A4

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Source: JBMO Shortlist 2023, A4

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#1 h Jun 28, 2024, 6:00 AM • 1 Y

Y by xytunghoanh

Let $a,b,c,d$ be positive real numbers with $abcd=1$ . Prove that

$\sqrt{\frac{a}{b+c+d^2+a^3}}+\sqrt{\frac{b}{c+d+a^2+b^3}}+\sqrt{\frac{c}{d+a+b^2+c^3}}+\sqrt{\frac{d}{a+b+c^2+d^3}} \leq 2$

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#3 h Jun 28, 2024, 9:42 AM • 2 Y

Y by aqusha_mlp12, xytunghoanh

By Cauchy-Schwarz or Jensen, we have
$\sum_{cyc}{\sqrt{\dfrac{a}{b+c+d^2+a^3}}}\leq \sqrt{4\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}}\leq 2$ It sufficies to show
$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq 1$ But by C-S $\left(b+c+d^2+a^3\right)\left(b+c+1+\dfrac{1}{a}\right)\geq \left(a+b+c+d\right)^2$ then
$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq \dfrac{\sum\limits_{cyc}{a\left(b+c+1+\dfrac{1}{a}\right)}}{\left(a+b+c+d\right)^2}$ $=\dfrac{\sum\limits_{sym}{(ab)}+ac+bd+a+b+c+d+4}{\left(a+b+c+d\right)^2}\overbrace{\leq}^{?} 1$ which is easy since
$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$ $\Longleftrightarrow a^2+b^2+c^2+d^2\geq a+b+c+d$ where the last inequality is true by $abcd=1$ .
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This post has been edited 1 time. Last edited by ehuseyinyigit, Aug 1, 2024, 10:58 PM

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#4 h Jun 28, 2024, 9:54 AM • 1 Y

Y by Assassino9931

I think that main idea in the problem is the same with JBMO 2019 #A.5. These problems are very related in their basis.

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3 posts

Duk168

#6 h Aug 21, 2024, 3:45 PM

Y by

ehuseyinyigit wrote:

By Cauchy-Schwarz or Jensen, we have
$\sum_{cyc}{\sqrt{\dfrac{a}{b+c+d^2+a^3}}}\leq \sqrt{4\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}}\leq 2$ It sufficies to show
$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq 1$ But by C-S $\left(b+c+d^2+a^3\right)\left(b+c+1+\dfrac{1}{a}\right)\geq \left(a+b+c+d\right)^2$ then
$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq \dfrac{\sum\limits_{cyc}{a\left(b+c+1+\dfrac{1}{a}\right)}}{\left(a+b+c+d\right)^2}$ $=\dfrac{\sum\limits_{sym}{(ab)}+ac+bd+a+b+c+d+4}{\left(a+b+c+d\right)^2}\overbrace{\leq}^{?} 1$ which is easy since
$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$ $\Longleftrightarrow a^2+b^2+c^2+d^2\geq a+b+c+d$ where the last inequality is true by $abcd=1$ .
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can you expl why
a^2 + b^2 + c^2 + d^2 >= a + b + c + d for abcd = 1

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arqady

#7 h Aug 21, 2024, 3:49 PM • 1 Y

Y by ehuseyinyigit

Duk168 wrote:

can you expl why
a^2 + b^2 + c^2 + d^2 >= a + b + c + d for abcd = 1

It's Muirhead

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89 posts

#8 h Nov 15, 2024, 2:08 PM

Y by

Duk168 wrote:

ehuseyinyigit wrote:

By Cauchy-Schwarz or Jensen, we have
$\sum_{cyc}{\sqrt{\dfrac{a}{b+c+d^2+a^3}}}\leq \sqrt{4\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}}\leq 2$ It sufficies to show
$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq 1$ But by C-S $\left(b+c+d^2+a^3\right)\left(b+c+1+\dfrac{1}{a}\right)\geq \left(a+b+c+d\right)^2$ then
$\sum_{cyc}{\dfrac{a}{b+c+d^2+a^3}}\leq \dfrac{\sum\limits_{cyc}{a\left(b+c+1+\dfrac{1}{a}\right)}}{\left(a+b+c+d\right)^2}$ $=\dfrac{\sum\limits_{sym}{(ab)}+ac+bd+a+b+c+d+4}{\left(a+b+c+d\right)^2}\overbrace{\leq}^{?} 1$ which is easy since
$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$ $\Longleftrightarrow a^2+b^2+c^2+d^2\geq a+b+c+d$ where the last inequality is true by $abcd=1$ .
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can you expl why
a^2 + b^2 + c^2 + d^2 >= a + b + c + d for abcd = 1

u can also use c.s like;
$(1+1+1+1)(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2$
$a^2+b^2+c^2+d^2 \ge \frac{(a+b+c+d)^2}{4}$
$(a+b+c+d)^2 \ge 4*(a+b+c+d)$ ==> $a+b+c+d \ge 4$ which is true by $AM-GM$

This post has been edited 2 times. Last edited by zaidova, Mar 4, 2025, 7:55 PM

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MR.1

#10 h May 9, 2025, 11:01 AM • 1 Y

Y by MIC38

first notice that $a^2+b^2+c^2+d^2\geq a+b+c+d$ from $abcd=1$
by Cauchy-Schwarz we have: $(b+c+d^2+a^3)(b+c+1+\frac{1}{a})\geq (a+b+c+d)^2$
so $\sqrt{\frac{a}{b+c+d^2+a^3}}\leq\sqrt{\frac{b+c+1+\frac{1}{a}}{(a+b+c+d)^2}}=\frac{\sqrt{b+c+a+\frac{1}{a}}}{a+b+c+d}$
so $LHS\leq \sum_{cyc}{\frac{\sqrt{b+c+1+\frac{1}{a}}}{a+b+c+d}}\leq \sqrt{{\sum_{cyc}{\frac{4(b+c+1+\frac{1}{a})}{(a+b+c+d)^2}}}}$
so we have to prove that $\sum_{sym}{b+c+1+\frac{1}{a}}\leq (a+b+c+d)^2\implies$
$4+ab+bc+cd+ad+2ac+2bd+(a+b+c+d)\leq (a+b+c+d)^2$
$4+ab+bc+cd+ad+2ac+2bd+(a+b+c+d)\leq a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$
$a^2+b^2+c^2+d^2+\left(a+c\right)\left(b+d\right)\geq a+b+c+d+4$
$a^2+b^2+c^2+d^2\geq a+b+c+d$ and we are done(i hate my life) :noo:

:noo:

:stretcher:

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