JBMO Shortlist 2023 A5

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Source: JBMO Shortlist 2023, A5

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#1 h Jun 28, 2024, 6:04 AM • 1 Y

Y by GeoKing

Let $a \geq b \geq 1 \geq c \geq 0$ be real numbers such that $a+b+c=3$ . Show that

$3 \left( \frac{a}{b}+\frac{b}{a} \right ) \geq 4c^2+\frac{a^2}{b}+\frac{b^2}{a}$

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#6 h Jun 28, 2024, 9:20 AM • 3 Y

Y by GeoKing, Math_.only., Novmath

The inequality is equilavent to
$3\left(a^2+b^2\right)\geq 4abc^2+a^3+b^3$ Since $a^3+b^3=\left(a+b\right)\left(a^2+b^2\right)-ab(a+b)$ , we have
$\Longleftrightarrow \left(3-a-b\right)\left(a^2+b^2\right)+ab(a+b)\geq 4abc^2$ by the problem condition, it boils down proving
$a^2c+b^2c+a^2b+ab^2\geq 4abc^2$ which is true by of course AM-GM
$a^2c+b^2c+a^2b+ab^2\geq 4ab\sqrt[4]{abc^2}\geq 4abc^2\Longleftrightarrow c\leq 1$ .

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1368 posts

#7 h Jul 1, 2024, 8:56 PM • 1 Y

Y by GeoKing

2 min solve by $s=a+b$ and $p=ab$ . (By the way, $a\geq b \geq 1$ is not needed, only $c\leq 1$ suffices.)

We have $2 \leq s < 3$ and want to prove $\frac{3(s^2-2p)}{p} \geq 4(3-s)^2 + \frac{s^3-3sp}{p}$ , equivalent to $\frac{s^2(3-s)}{p} \geq 4s^2 - 27s + 42$ . Since $p \leq \frac{s^2}{4}$ (equivalent to $(a-b)^2 \geq 0$ ), it suffices to have $4(3-s) \geq 4s^2-27s+42$ . This is equivalent to $4s^2 - 23s + 30 \leq 0$ , i.e. $(s-2)(2s-15) \leq 0$ , which holds for the given interval for $s$ .

This post has been edited 1 time. Last edited by Assassino9931, Jul 1, 2024, 8:57 PM

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Wild

#8 h Jul 5, 2024, 5:36 PM • 2 Y

Y by GeoKing, zaidova

$(a+b+c)(a/b+b/a)>=4c^2+a^2/b+b^2/a$
$a^2/b+b+a+b^2/a+ac/b+bc/a>=4c^2+a^2/b+b^2/a$
$b+a+bc/a+ac/b>=4c^2$
$b>=1>=c>=c^2$
$a>=1>=c>=c^2$
the rest is simple

This post has been edited 4 times. Last edited by Wild, Jul 5, 2024, 5:44 PM

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#9 h Jul 16, 2024, 11:09 AM

Y by

Solution(I hope this is true):
Multiply both sides with ab and we get 3(a^2+b^2)=>4abc^2+a^3+b^3.Because a+b+c=3,the inequality can be rewritten as ba^2+ab^2+c(a^2+b^2)=>4abc^2.Now,observe that a+b=>2c and ab^2+ba^2=ab(a+b)=>2abc and the inequality can be rewritten as (a+b)^2=>4cab which is true because c<=1 and we are done.

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#11 h Nov 11, 2024, 3:16 PM

Y by

$3=a+b+c$ we will replace it in the given inequality.
$3*(\frac{a}{b}+\frac{b}{a})=(a+b+c)(\frac{a}{b}+\frac{b}{a})\ge 4*c^2+\frac{a^2}{b}+\frac{b^2}{a}$
Then inequality will be;
$(a+b+c)(a^2+b^2) \ge(4*c^2*a*b+a^3+b^3)$
then just simplifize;
$a*(b^2)+(a^2)*b+c*(a^2)+c*(b^2) \ge 4c^2*a*b$
<==> $ab*(a+b)+c*(a^2+b^2) \ge ab*(a+b)+ 2*a*b*c \ge 4c^2*a*b$
divide ab;
$a+b+2c \ge 4*c^2$ $(1)$
$a \ge c$ , $b \ge c$ <==> $a+b \ge 2*c$
from $(1)$ ;
$4*c \ge 4*c^2$ <==> $1 \ge c$ which is true (given condition)

This post has been edited 1 time. Last edited by zaidova, Nov 11, 2024, 3:26 PM

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#12 h Dec 22, 2024, 8:06 PM

Y by

Clearing out the denominators, the inequality becomes

$3(a^2 + b^2) \geq 4abc^2 + a^3 + b^3$
Note that $a^3 + b^3 = (a+b)(a^2 + b^2) - a^2b - ab^2$ . Using this, the inequality becomes

$3(a^2 + b^2) \geq 4abc^2 + (a+b)(a^2 + b^2) - a^2b - ab^2$ $(3 - a - b)(a^2 + b^2) + ab(a + b) \geq 4abc^2$ $c(a^2 + b^2) + ab(3 - c) \geq 4abc^2$
By AM-GM, we can write $a^2 + b^2 \geq 2ab$ .

$c(2ab) + 3ab - abc \geq 4abc^2$ $2c + 3 - c \geq 4c^2$ $0 \geq 4c^2 - c - 3 = (4c + 3)(c - 1)$
Because $c$ is in the interval $[0, 1]$ , $(4c + 3)(c - 1)$ will be negative, as desired.

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#13 h Feb 23, 2025, 6:53 PM • 1 Y

Y by definite_denny

Rewrite as $a(\frac{3-a}{b})+b(\frac{3-b}{a}) \geq 4c^2$ and $AM-GM$ on the expression in LHS gives $\sqrt{(b+c)(a+c)} \geq 2c^2$ .Write CS in the square root to get $\sqrt{ab}+c \geq 2c^2,\Rightarrow \sqrt{ab} \geq c(2c-1)$ .Just seperate into cases $1)0<c<0,5$
$2)c=0;0,5$
$3)0,5,<c<1$
$4)c=1$ and finish the problem.

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#14 h Apr 18, 2025, 11:01 AM

Y by

After clearing denominators ,add $3c^2$ + $3c^3$ to both sides and replace 3 with a+b+c to get:

$a^2c+b^2c+a^2b+ab^2\geq 4abc^2$
which is true by substituting one of the “c”s as 1 and use some AM-GM or just by directly using AM-GM to whole ineq.

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sqing

#15 h Apr 18, 2025, 2:11 PM

Y by

Let $a \geq b \geq 1 \geq c \geq 0$ and $a+b+c=3$ . Show that
$6\left( \frac{a}{b}+\frac{b}{a}-c\right ) \geq 4c^2+\frac{a^2}{b}+\frac{b^2}{a}$

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Primeniyazidayi

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Primeniyazidayi

#16 h Apr 18, 2025, 3:35 PM

Y by

sqing wrote:

Let $a \geq b \geq 1 \geq c \geq 0$ and $a+b+c=3$ . Show that
$6\left( \frac{a}{b}+\frac{b}{a}-c\right ) \geq 4c^2+\frac{a^2}{b}+\frac{b^2}{a}$

A sketch for solution.

It succifies to show that $6(a^2+b^2-abc) \geq 4abc^2+a^3+b^3$ .Note that $a^3+b^3=(3-c)(c^2-6c+9-3ab)=(3-c)^3-(9-3c)ab$ .After some algebraic manipulations,it is enough to show that $ab \geq \frac{c^3-3c^2-9c+27}{4c^2+9c+3}$ But it can be shown that $f(c)=\frac{c^3-3c^2-9c+27}{4c^2+9c+3} \leq 1$ in the interval $[1,3)$ ,done.

This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 18, 2025, 3:35 PM

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sqing

#17 h Apr 18, 2025, 4:00 PM

Y by

Thanks.

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