Guess who is the author?

mihaig wrote: Guess who is the author? I’m not good at guessing , so... is it you dear sir?

DoThinh2001 wrote: mihaig wrote: Guess who is the author? I’m not good at guessing , so... is it you dear sir? Later.

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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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

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

ali3985 wrote: Bmo 2019 problems https://m.facebook.com/story.php?story_fbid=1265546146937548&id=568380743320762 Thank you very much.

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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.

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

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.

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.

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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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

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 $\sqrt{bc}(a+1)\ge 2\sqrt{abc} \ge 2.$

rmtf1111 wrote: In my opinion, the following solution is the only one that makes the problem suitable for an olympiad. Case 1: $b\ge 1 \implies c\ge 1$ $$\sqrt{bc}(a+1)\ge \frac{2bc(a+1)}{b+c}=2\left(1+\frac{a(b-1)(c-1)}{b+c}\right) \ge 2$$Case 2: $b\le 1,$ then it is enough to consider $c>1,$ 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 $\sqrt{bc}(a+1)\ge 2\sqrt{abc} \ge 2.$ I also did solve in this way.

Your opinion has been noted. Are you some president of the competition?

mihaig wrote: Are you some president of the competition? No. However, I attached some photos so you can see who are the presidents in case it interests you

mihaig wrote: Your opinion has been noted. Are you some president of the competition? I think this is very good problem, and I see a lot of beautiful problems of yours here, also one problem of yours has been in TST of Kosovo which I appretiate, but I don't think you have to be rude at him, I mean it is his opinion and everybody should rescpect each other opinions as long as don't offend anyone. So please be more open mind and polite I tell you as a friend.

Dear @author of this problem everyone have a opinion and being rude is not good

hellomath010118 wrote: bel.jad5 wrote: hellomath010118 wrote: Here is a nice solution: Since $(a+b+c)^2 \geq 3(ab+bc+ca)=3(a+b+c)$ and $a+b+c>0 \implies a+b+c \geq 3$. Since $c$ is the maximum of the three numbers thus, $c\geq1$. But $a+b+c=ab+bc+ca \implies c=\dfrac{ab-a-b}{1-a-b} \geq 1$ and we get $ab \geq 1$. Hence $abc\geq 1$ and $\sqrt{bc} (a+1) \geq \sqrt{bc} \cdot 2 \sqrt{a}\geq2. \blacksquare$ 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) Javie4284 wrote: bel.jad5 wrote: why $ab\geq 1$? We don't need to take mister helios seriously. The recent history proved us that he is inventing a lot of balast. But nothing good. Can you tell me how?( I am not being rude .) $ab\geq 1$ is not always true...$(0,2,2)$ for example is an equality case for this problem.

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

hellomath010118 wrote: 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 $a=10^{-2019}$

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 $ab \geq 1$

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

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, $\sqrt{bc}(a+1)\geq 2. \blacksquare$

Bravo! Interesting perspective.

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

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?

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 $ab=p^2?$

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 $c\geq b\geq a\geq0~\text{and}~ab+bc+ac=a+b+c>0~.$

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 $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$

Marius Stannean Wrong solution Aldab ketgansan

Can you detail your assertion?

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: $$ (y - 1)^2(2 - y) \geq 0, \text{ and we are done.} $$