Inequality with condition a+b+c = ab+bc+ca (and special equality case)

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DoThinh2001

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DoThinh2001

#1 h May 2, 2019, 11:41 AM • 7 Y

Y by mihaig, dangerousliri, Pluto1708, rightways, jhu08, Adventure10, Mango247

Let $a,b,c$ be real numbers such that $0 \leq a \leq b \leq c$ and $a+b+c=ab+bc+ca >0.$
Prove that $\sqrt{bc}(a+1) \geq 2$ and determine the equality cases.

(Edit: Proposed by sir Leonard Giugiuc, Romania)

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mihaig

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mihaig

#2 h May 2, 2019, 11:48 AM • 1 Y

Y by Adventure10

Guess who is the author?

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DoThinh2001

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DoThinh2001

#3 h May 2, 2019, 12:07 PM • 5 Y

Y by mihaig, teomihai, Adventure10, Mango247, Mango247

mihaig wrote:

Guess who is the author?

I’m not good at guessing

, so... is it you dear sir?

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mihaig

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#4 h May 2, 2019, 12:13 PM • 1 Y

Y by Adventure10

DoThinh2001 wrote:

mihaig wrote:

Guess who is the author?

I’m not good at guessing

, so... is it you dear sir?

Later.

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sqing

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sqing

#7 h May 2, 2019, 1:07 PM • 2 Y

Y by Adventure10, Nari_Tom

DoThinh2001 wrote:

Let $a,b,c$ be real numbers such that $0 \leq a \leq b \leq c$ and $a+b+c=ab+bc+ca >0.$
Prove that $\sqrt{bc}(a+1) \geq 2$ and determine the equality cases.

ali3985 wrote:

Clearly $a\le 1, c \ge 1$ , since

$a+b+c \le c(a+b+c)$

and

$a+b+c \ge a(a+b+c)$

Then

$(b+c)(1-a)=bc-a \ge 2\sqrt{bc}(1-a)$

$\iff \sqrt{bc} \ge \sqrt{(1-a)^2+a}+1-a \ge \frac{2}{1+a} \iff a(a-1)^2\ge 0$

See here related problem
https://artofproblemsolving.com/community/q2h1825456p12213876

Proof of Wujiangzheng (Tsinghua University)：

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mihaig

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#8 h May 2, 2019, 1:15 PM • 2 Y

Y by Adventure10, Mango247

No,no,no,no,no .
See here the main problem that generated and inspired that author.
Not #7 or that very few points solution.
https://artofproblemsolving.com/community/c6h1713418p12187873

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ali3985

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ali3985

#9 h May 2, 2019, 1:25 PM • 2 Y

Y by Adventure10, Mango247

Bmo 2019 problems
https://m.facebook.com/story.php?story_fbid=1265546146937548&id=568380743320762

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sqing

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sqing

#10 h May 2, 2019, 1:35 PM • 2 Y

Y by ali3985, Adventure10

ali3985 wrote:

Bmo 2019 problems
https://m.facebook.com/story.php?story_fbid=1265546146937548&id=568380743320762

Thank you very much.

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mihaig

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#11 h May 2, 2019, 2:18 PM • 14 Y

Y by ali3985, DoThinh2001, Golub_Srecko, MathPassionForever, Pluto1708, HECAM-CA-CEBEPA, CP31, sahi99, enzoP14, Bartina, Aryan-23, Elyson, Adventure10, Zehra19

DoThinh2001 wrote:

Let $a,b,c$ be real numbers such that $0 \leq a \leq b \leq c$ and $a+b+c=ab+bc+ca >0.$
Prove that $\sqrt{bc}(a+1) \geq 2$ and determine the equality cases.

See here my solution.
I didn't see the official solution.
Let $a+b+c=ab+bc+ca=k$ . Since $(a+b+c)^{2}\ge3(ab+bc+ca)$ ,
we get that $k^{2}\ge3k$ . Because $k>0$ , we obtain that $k\ge3$ .

We have $bc\ge ca\ge ab$ , so from the above relation we deduce that
$bc\ge1$ .

Denote $b+c=2s$ and $bc=p^{2}$ . By AM-GM, $s\ge p$ . Furthermore,
$s=p$ if and only if $b=c$ . Also, $p\ge1$ .

Case 1: $1\le p\le2$ .

The constraint gives us $a=\frac{b+c-bc}{b+c-1}=\frac{2s-p^{2}}{2s-1}$ .

For any $t\ge p$ , we have $\frac{2t-p^{2}}{2t-1}=1-\frac{p^{2}-1}{2t-1}$ . Hence, $a=\frac{2s-p^{2}}{2s-1}\ge\frac{2p-p^{2}}{2p-1}=\frac{p}{2p-1}\cdot(2-p)$

$=>pa\ge\frac{p^{2}}{2p-1}\cdot(2-p)$ .

But $\frac{p^{2}}{2p-1}\ge1=>pa\ge2-p=>p(a+1)\ge2=>\sqrt{bc}(a+1)\ge2$ .

The equality holds when $b=c$ and $p=1$ or $p=2$ , i.e. $a=b=c=1$
or $a=0$ and $b=c=2$ .

Case 2: $p>2$ .

We have $\sqrt{bc}(a+1)=p(a+1)>2(a+1)\ge2$ .

The proof is complete.

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MathPassionForever

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MathPassionForever

#12 h May 2, 2019, 3:37 PM • 1 Y

Y by Adventure10

For the first time I solved an olympiad inequality, and I got the exact same procedure as @above.

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MariusStanean

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MariusStanean

#13 h May 2, 2019, 4:02 PM • 12 Y

Y by Pluto1708, rightways, Supermathlet_04, HECAM-CA-CEBEPA, enzoP14, Bartina, Infinityfun, Creeper1612, ytChen, Adventure10, Mango247, ant_

DoThinh2001 wrote:

Let $a,b,c$ be real numbers such that $0 \leq a \leq b \leq c$ and $a+b+c=ab+bc+ca >0.$
Prove that $\sqrt{bc}(a+1) \geq 2$ and determine the equality cases.

1. If $bc\ge 4$ done
2. Let $bc\le 4$ then denoting $t=\sqrt{bc}\ge 1$ , since $3bc\ge ab+bc+ca=\frac{(a+b+c)^2}{ab+bc+ca}\ge 3$ , we have
$\sqrt{bc}(a+1) \geq 2\Longleftrightarrow t\left(\frac{2(b+c)-bc-1}{b+c-1}\right)\ge 2\Longleftrightarrow 2(b+c)(t-1)-t^3-t+2\ge0$ but $b+c\ge 2t$ so it remains to prove that
$t^3-4t^2+5t-2\le0\Longleftrightarrow (t-1)^2(t-2)\le0$ true.

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mihaig

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mihaig

#14 h May 2, 2019, 4:39 PM • 2 Y

Y by MathPassionForever, Adventure10

MathPassionForever wrote:

For the first time I solved an olympiad inequality, and I got the exact same procedure as @above.

Also did I .This is a beautiful coincidence,or rather a miracle.
Which is extraordinary.

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mihaig

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mihaig

#15 h May 2, 2019, 4:58 PM • 2 Y

Y by Adventure10, Mango247

DoThinh2001 wrote:

Let $a,b,c$ be real numbers such that $0 \leq a \leq b \leq c$ and $a+b+c=ab+bc+ca >0.$
Prove that $\sqrt{bc}(a+1) \geq 2$ and determine the equality cases.

Apart the beauty of the problem,see below it's relevance as Theorem.
I used it back in 2016-2017 for my Crux 4300 from the link below.
See the marked part.
P.S. This is not equivalent to "the problem is old".It is new actually,and never published before.Bu if we knew how and where to look...More importantly,how.
https://cms.math.ca/crux/v43/n10/Problems_43_10.pdf

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#16 h May 2, 2019, 6:25 PM • 1 Y

Y by Adventure10

See here new approaches from Greece.
Note that Demetres is the author of Problem 4.
https://mathematica.gr/forum/viewtopic.php?f=58&t=64382&p=311464#p311464

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rmtf1111

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rmtf1111

#17 h May 2, 2019, 7:06 PM • 10 Y

Y by dangerousliri, abdelkrim, enzoP14, Kobayashi, Bartina, Aryan-23, MelonGirl, Adventure10, Mango247, Tellocan

In my opinion, the following solution is the only one that makes the problem suitable for an olympiad.
Case 1: $c\ge b\ge 1$ or $b\le c\le 1$
$\sqrt{bc}(a+1)\ge \frac{2bc(a+1)}{b+c}=2\left(1+\frac{a(b-1)(c-1)}{b+c}\right) \ge 2$
Case 2: $b\le 1$ and $c>1,$ note that $abc-1=(a-1)(b-1)(c-1),$ and since $a-1\le b-1\le 0<c-1,$ we obtain that $abc>1.$ Now apply AM-GM $\sqrt{bc}(a+1)\ge 2\sqrt{abc} \ge 2.$

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bel.jad5

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bel.jad5

#60 h Dec 22, 2019, 4:05 PM • 1 Y

Y by Adventure10

hellomath010118 wrote:

bel.jad5 wrote:

why $ab\geq 1$ ?

I solved for $c$ using $a+b+c=ab+bc+ca$ to obtain $(ab-a-b)/(1-a-b)=c\geq 1$ .I should have written it the other way round for more clarity. Edit: It might seem that this step is invalid for $a+b=1$ but then $a+b+c=ab+bc+ca \implies ab=1$ which is impossible for $a+b=1$ (due to AM-GM)

bel.jad5 wrote:

$ab\geq 1$ is not always true... $(0,2,2)$ for example is an equality case for this problem.

Then it must be true for $a \neq 0$ and the case $a=0$ is easy. Sorry, my silly fault of not considering this case

I think it is still not true for $a$ very small. take $a=10^{-2019}$

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hellomath010118

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hellomath010118

#61 h Dec 23, 2019, 9:16 AM • 3 Y

Y by math_comb01, Adventure10, Mango247

I'm afraid that you are true ; the hole is in this step I think:

hellomath010118 wrote:

$a+b+c=ab+bc+ca \implies \dfrac{ab-a-b}{1-a-b}=c \geq 1$ and we get $ab \geq 1$

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mihaig

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#62 h Dec 23, 2019, 10:53 AM • 1 Y

Y by Adventure10

Don't worry, with a little patience you'll find an own interesting proof.

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Keith50

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Keith50

#64 h Aug 26, 2020, 8:28 AM • 1 Y

Y by Iora

Since there are many synthetic proofs, let me present a proof using Lagrange Multipliers.
Define $f(a,b,c)=bc(a^2+2a+1)$ and $g(a,b,c)=ab+bc+ca-a-b-c$ , so that we will prove that $f(a,b,c)\geq 4$ with $g(a,b,c)=0$ using Lagrange Multipliers. Note that $\nabla g=\left<b+c-1,c+a-1,a+b-1\right>\ne 0$ at all points, and moreover that $f$ and $g$ have continuous partial derivatives.

Let $U=\{(a,b,c) | a^2+b^2+c^2 <1000\}$ , then $\overline U=\{(a,b,c) | a^2+b^2+c^2\leq 1000\}$ which is bounded, then the constraint set, $\overline S=\{\mathbf x \in \overline U : g(\mathbf x)=0\}$ is compact. Hence, it achieves a minimum value say $\mathbf x$ .

Let $\mathbf x$ lies on the boundary, then at least one component of $\mathbf x$ is zero, let's say $a=0$ , then we have $bc=b+c \implies \frac{b+c}{2}\geq \sqrt{bc}, bc\geq 4 \iff f(0,b,c)=bc\geq 4$ as desired and the minimum value holds at $a=0, b=c=2$ .

Otherwise, let $\mathbf x=(a,b,c)$ . By Lagrange Multiplies, we have $\left<bc(2a+2), c(a^2+2a+1), b(a^2+2a+1)\right>=\lambda \left<b+c-1, c+a-1, a+b-1\right>.$ If $\lambda =0,$ then $b=0$ , $c=0$ which don't satisfy the conditions that $ab+bc+ca >0$ . So, $\lambda \ne 0$ , we will get that $2bc(a+1)=\lambda (b+c-1) \ \ (1)$ $c(a+1)^2=\lambda (c+a-1) \ \ (2)$ $b(a+1)^2=\lambda (a+b-1) \ \ (3)$ From $(1), \lambda=\frac{2bc(a+1)}{b+c-1}$ , sub it into $(2)$ , we get $a=-1+\frac{4b-2bc}{b-c+1}$ and so from $(3)$ , $b(a+1)^2=\frac{2bc(a+1)(b+a-1)}{b+c-1}\implies a=-1+\frac{2bc-4c}{b-c-1}.$ Equating these two equalities, $\frac{4b-2bc}{b-c+1}=\frac{2bc-4c}{b-c-1}\implies (b-c)(b+c-bc-1)=0$

Case 1: $b=c$
We get $a=-1+\frac{2bc-4c}{b-c-1}=-2c^2+4c-1$ and from $ab+bc+ca=a+b+c,$ $c(-2c^2+4c-1)+c^2+c(-2c^2+4c-1)-(-2c^2+4c-1)-c-c=0 \implies (c-1)(4c^2-7c+1)=0$ If $c=b=1$ , then $a=1$ and $f(1,1,1)=4$ which means $\mathbf x=(1,1,1)$ . If $4c^2-7c+1=0, c=\frac{7\pm \sqrt{33}}{8}$ and solving for $a$ gives $a=\frac{-1\pm \sqrt{33}}{16}$ . Take $a=\frac{-1+\sqrt{33}}{16}\geq 0$ , then we can check that $f\left(\frac{-1+\sqrt{33}}{16}, \frac{7+\sqrt{33}}{8}, \frac{7+\sqrt{33}}{8}\right)\geq 4.$ Case 2: $b+c-bc-1=0$
We have $(b-1)(c-1)=0$ \implies $b=1$ or $c=1$ . In either case, we can get that $\mathbf x=(1,1,1)$ also.
HENCE, the equality holds when $(a,b,c)=(0,2,2), (1,1,1)$ . Since this is the minimum point, $\sqrt{bc}(a+1)\geq 2. \blacksquare$

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mihaig

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#65 h Aug 28, 2020, 8:03 PM • 3 Y

Y by Mango247, Mango247, Mango247

Bravo! Interesting perspective.

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duongbgbg32

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duongbgbg32

#66 h Oct 16, 2020, 12:47 PM • 1 Y

Y by Wizard0001

My solution
Let $a+b+c=ab+bc+ac=k$ . We have $(a+b+c)^2 \ge 3(ab+bc+ca)$ So $k^2 \ge 3k \implies k\ge 3$ .
From the question, we have $bc \ge ca \ge ab$ . So $bc\ge 1$ .
We have $b+c \ge 2\sqrt{bc} \ge 2$
On the other hand, we have $a=\frac{b+c-bc}{b+c-1}=1-\frac{bc-1}{b+c-1} \ge 1-\frac{bc-1}{2\sqrt{bc}-1}=\frac{\sqrt{bc}(2-\sqrt{bc})}{2\sqrt{bc}-1}$ For $\sqrt{bc}=2$ , we have $a\ge 0$ . So $\sqrt{bc}(a+1) \ge 2$ For $\sqrt{bc} >2$ , $\sqrt{bc}(a+1)>2(a+1) \ge 2$ For $\sqrt{bc}<2$ , we have $a\sqrt{bc}+\sqrt{bc} \ge \frac{\sqrt{bc}(2-\sqrt{bc})}{2\sqrt{bc}-1}+\sqrt{bc}=\frac{bc}{2\sqrt{bc}-1}(2-\sqrt{bc})+\sqrt{bc}\ge 2-\sqrt{bc}+\sqrt{bc}=2$ Q.E.D

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mihaig

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#67 h Oct 16, 2020, 3:39 PM

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Bravo! https://artofproblemsolving.com/community/c6t243f6h2305712_non_symmetric_with_3suma2sumabgt0

Let $a\geq b\geq c\geq d\geq0$ such that $3\left(a+b+c+d\right)=2\left(ab+bc+cd+da+ac+bd\right)>0.$ Prove
$\sqrt{ab}+\sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}+\sqrt{cd}+a-d\geq6.$ When do we have equality?

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mihaig

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mihaig

#68 h Oct 20, 2020, 7:07 AM • 3 Y

Y by Mango247, Mango247, Mango247

For the collection, another own problem.

For what positive real values of $p$ does
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\geq3$ hold for all $a\geq b\geq c\geq0$ satisfying $a+b+c=ab+bc+ca>0$ and $ab=p^2?$

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mihaig

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mihaig

#69 h Jun 24, 2021, 5:03 AM

Y by

For the collection: https://artofproblemsolving.com/community/c6h2599598p22438129

Let $k>0~$ be fixed. Find
$\max\left(\frac{1}{ab+k}+\frac{1}{bc+k}+\frac{1}{ac+k}\right)$ over all $c\geq b\geq a\geq0~\text{and}~ab+bc+ac=a+b+c>0~.$

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mihaig

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#70 h Jul 14, 2021, 2:52 AM

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For the collection: https://artofproblemsolving.com/community/c6h1740846p22510809

Let $k>0~$ be fixed. Find
$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$

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IMO2022Goldinshallah

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IMO2022Goldinshallah

#71 h Aug 6, 2021, 4:49 AM

Y by

Marius Stannean
Wrong solution
Aldab ketgansan

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mihaig

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#72 h Aug 12, 2021, 1:02 PM

Y by

Can you detail your assertion?

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Nari_Tom

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#73 h Jan 19, 2025, 9:09 AM

Y by

If $\sqrt{bc} > 2$ , then we have $\sqrt{bc}(a+1) > 2$ .
If $\sqrt{bc} < 1$ , this implies $bc < 1$ which is $ab + ac + bc < 3$ , impossible.
When $1 \leq \sqrt{bc} \leq 2$ , let's denote $x = b + c$ and $y = \sqrt{bc}$ .
Then we have:
$(a+1)\sqrt{bc} = \left( 2b + 2c - bc - \frac{1}{b+c-1} \right)\sqrt{bc} \geq 2.$
This is equivalent to proving:
$2x(y - 1) \geq y^3 + y - 2.$
If we use $x \geq 2y$ , it factorizes to:
$(y - 1)^2(2 - y) \geq 0, \text{ and we are done.}$

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Rayvhs

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Rayvhs

#75 h Apr 27, 2025, 7:00 PM

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$(ab+bc+ac)^2 = (a+b+c)^2 \geq 3(ab+bc+ac)$
$\iff ab+bc+ac \geq 3 \Rightarrow a+b+c \geq 3, \quad a \leq b \leq c \quad \Rightarrow \quad c \geq 1.$
$c = \frac{ab - a - b}{1 - a - b} \geq 1 \quad \Rightarrow \quad abc \geq 1.$

Now,
$\sqrt{bc}(a+1) \geq \sqrt{\frac{1}{a}}(a+1) \geq \sqrt{\frac{1}{a}} \times 2\sqrt{a} = 2.$

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mihaig

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mihaig

#76 h Apr 30, 2025, 1:58 PM

Y by

Rayvhs wrote:

$(ab+bc+ac)^2 = (a+b+c)^2 \geq 3(ab+bc+ac)$
$\iff ab+bc+ac \geq 3 \Rightarrow a+b+c \geq 3, \quad a \leq b \leq c \quad \Rightarrow \quad c \geq 1.$
$c = \frac{ab - a - b}{1 - a - b} \geq 1 \quad \Rightarrow \quad abc \geq 1.$

Now,
$\sqrt{bc}(a+1) \geq \sqrt{\frac{1}{a}}(a+1) \geq \sqrt{\frac{1}{a}} \times 2\sqrt{a} = 2.$

How $abc\geq1?$ It's false, of course

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