Something nice

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KhuongTrang

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KhuongTrang

#1 h Nov 1, 2023, 12:56 PM

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Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$

This post has been edited 2 times. Last edited by KhuongTrang, Nov 19, 2023, 11:59 PM

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mihaig

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mihaig

#2 h Nov 1, 2023, 3:42 PM

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Beauty. But difficult

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KhuongTrang

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KhuongTrang

#19 h Nov 9, 2023, 1:09 PM • 1 Y

Y by MihaiT

Non sense post.

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KhuongTrang

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KhuongTrang

#31 h Nov 19, 2023, 5:34 AM

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Something not relevant

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arqady

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arqady

#32 h Nov 19, 2023, 6:24 AM • 1 Y

Y by teomihai

KhuongTrang wrote:

Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that $a\sqrt{bc+1}+b\sqrt{ca+1}+c\sqrt{ab+1}\ge 2\sqrt{a+b+c-1}.$

Because $\sum_{cyc}a\sqrt{bc+1}=\sqrt{\sum_{cyc}(a^2bc+a^2+2ab\sqrt{(bc+1)(ac+1)}}\geq\sqrt{\sum_{cyc}(a^2+2ab)}=a+b+c\geq2\sqrt{a+b+c-1}.$

This post has been edited 1 time. Last edited by arqady, Nov 19, 2023, 6:25 AM

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KhuongTrang

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KhuongTrang

#34 h Nov 20, 2023, 11:13 AM

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Something not relevant

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sqing

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sqing

#35 h Nov 20, 2023, 12:43 PM

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Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that $\sqrt{a+b+abc}+\sqrt{b+c+abc}+\sqrt{c+a+abc}\ge 2+\sqrt{2}$

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KhuongTrang

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KhuongTrang

#43 h Nov 30, 2023, 1:29 PM

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Something not relevant

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KhuongTrang

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#59 h Jun 12, 2024, 1:29 PM

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Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that
$\color{blue}{\sqrt{\frac{a+bc}{a+1}}+\sqrt{\frac{b+ca}{b+1}}+\sqrt{\frac{c+ab}{c+1}}\le 1+\sqrt{2}. }$

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mudok

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mudok

#60 h Jun 12, 2024, 3:00 PM • 1 Y

Y by arqady

arqady wrote:

KhuongTrang wrote:

Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that $a\sqrt{bc+1}+b\sqrt{ca+1}+c\sqrt{ab+1}\ge 2\sqrt{a+b+c-1}.$

Because $\sum_{cyc}a\sqrt{bc+1}=\sqrt{\sum_{cyc}(a^2bc+a^2+2ab\sqrt{(bc+1)(ac+1)}}\geq\sqrt{\sum_{cyc}(a^2+2ab)}=a+b+c\geq2\sqrt{a+b+c-1}.$

We can directly use: $\sum a\sqrt{bc+1}\ge \sum a$ :lol:

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KhuongTrang

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KhuongTrang

#65 h Jun 22, 2024, 3:22 AM

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Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=2.$ Prove that

$\color{blue}{\sqrt{15a+1} +\sqrt{15b+1} +\sqrt{15c+1}\ge 3\sqrt{3}\cdot\sqrt{1+2(ab+bc+ca)}. }$

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arqady

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arqady

#66 h Jun 22, 2024, 6:35 AM

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KhuongTrang wrote:

Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=2.$ Prove that

$\color{blue}{\sqrt{15a+1} +\sqrt{15b+1} +\sqrt{15c+1}\ge 3\sqrt{3}\cdot\sqrt{1+2(ab+bc+ca)}. }$

Holder with $(3a+1)^3$ and $uvw$ .

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KhuongTrang

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KhuongTrang

#72 h Jul 15, 2024, 1:47 AM • 1 Y

Y by ehuseyinyigit

KhuongTrang wrote:

Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that
$\color{blue}{\sqrt{\frac{a+bc}{a+1}}+\sqrt{\frac{b+ca}{b+1}}+\sqrt{\frac{c+ab}{c+1}}\le 1+\sqrt{2}. }$

Problem. Given non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that
$\color{blue}{\sqrt{\frac{a+b}{c+1}}+\sqrt{\frac{c+b}{a+1}}+\sqrt{\frac{a+c}{b+1}}\le 2\sqrt{a+b+c}. }$ Equality holds iff $a=b=1,c=0$ or $a=b\rightarrow 0,c\rightarrow +\infty$ and any cyclic permutations.

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arqady

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arqady

#73 h Jul 15, 2024, 4:50 PM

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KhuongTrang wrote:

Problem. Given non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that
$\color{blue}{\sqrt{\frac{a+b}{c+1}}+\sqrt{\frac{c+b}{a+1}}+\sqrt{\frac{a+c}{b+1}}\le 2\sqrt{a+b+c}. }$ Equality holds iff $a=b=1,c=0$ or $a=b\rightarrow 0,c\rightarrow +\infty$ and any cyclic permutations.

Because $\sum_{cyc}\sqrt{\frac{a+b}{c+1}}\leq\sqrt{\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{c+1}}\leq2\sqrt{a+b+c}.$ :-D

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bellahuangcat

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bellahuangcat

#74 h Jul 15, 2024, 4:54 PM

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KhuongTrang wrote:

Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$

what why does that look so easy and difficult at the same time lol

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ehuseyinyigit

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ehuseyinyigit

#75 h Jul 15, 2024, 5:38 PM

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That's the beauty of it

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bellahuangcat

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bellahuangcat

#76 h Jul 15, 2024, 6:01 PM

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ehuseyinyigit wrote:

That's the beauty of it

yeah ig

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arqady

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#78 h Jul 16, 2024, 7:35 AM

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sqing wrote:

Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that $\sqrt{a+b+abc}+\sqrt{b+c+abc}+\sqrt{c+a+abc}\ge 2+\sqrt{2}$

The following inequality is also true.
Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=1$ . Prove that:
$\sqrt{a+b+\frac{13}{14}abc}+\sqrt{b+c+\frac{13}{14}abc}+\sqrt{c+a+\frac{13}{14}abc}\ge 2+\sqrt{2}$

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KhuongTrang

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KhuongTrang

#83 h Aug 10, 2024, 3:56 PM

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Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=3.$ Prove that

$\color{blue}{\sqrt{\frac{4}{3}(ab+bc+ca)+5}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}.}$

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kiyoras_2001

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kiyoras_2001

#84 h Aug 10, 2024, 5:59 PM

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KhuongTrang wrote:

Problem. Given $a,b,c$ be non-negative real numbers such that $a+b+c=3.$ Prove that
$\color{blue}{\sqrt{\frac{4}{3}(ab+bc+ca)+5}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}.}$

After homogenizing and squaring it becomes
$\sum a^2+8\sum ab\ge 3\sum a\sum\sqrt{ab}.$ Changing $a\to a^2, b\to b^2, c\to c^2$ it becomes a fourth degree inequality, so is linear in $w^3$ . Thus it remains to check only the cases $c=0$ and $b=c=1$ which is easy.

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KhuongTrang

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#92 h Mar 26, 2025, 1:00 AM

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Problem. Let $a,b,c$ be non-negative real variables with $ab+bc+ca>0.$ Prove that $\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$ Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$

This post has been edited 1 time. Last edited by KhuongTrang, Mar 28, 2025, 1:21 AM

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jokehim

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#93 h Mar 26, 2025, 1:11 PM

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KhuongTrang wrote:

Problem. Let $a,b,c$ be non-negative real variables with $a+b+c>0.$ Prove that $\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$ Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$

Assume that $a+b+c=1$ and set $M=a^2b+b^2c+c^2a,\ \ ab+bc+ca=q,\ \ abc=r.$ The inequality becomes $10 M^2 - 16 M q + 12 M r - 8 q^3 + 8 q^2 - 51 q r + 63 r^2 + 10 r\ge 0$ ưhere $\Delta_M=8 (40 q^3 - 8 q^2 + 207 q r - r (297 r + 50))<0$ which ends the proofs

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KhuongTrang

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#94 h Mar 27, 2025, 4:23 AM

Y by

#93 Could you please check your solution again, jokehim? I think this inequality is very hard to think of a proof in normal way.
Hope to see some ideas. Btw, it is obviously true by BW.

This post has been edited 2 times. Last edited by KhuongTrang, Yesterday at 12:05 AM

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jokehim

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#95 h Mar 27, 2025, 6:23 AM

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KhuongTrang wrote:

#93 Could you please check your solution again, jokehim? I think this inequality is very hard to think of a proof in normal way.
Hope to see some ideas. Btw, it is obviously true by BW.

Problem. Let $a,b,c$ be positive real variables with $a+b+c+2\sqrt{abc}=1.$ Prove that $\frac{\sqrt{a+ab+b}}{\sqrt{ab}+\sqrt{c}}+\frac{\sqrt{b+bc+c}}{\sqrt{bc}+\sqrt{a}}+\frac{\sqrt{c+ca+a}}{\sqrt{ca}+\sqrt{b}}\ge 3.$ Equality holds iff $a=b=c=\dfrac{1}{4}.$

I don't see what's wrong with my solution

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Nguyenhuyen_AG

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Nguyenhuyen_AG

#96 h Mar 27, 2025, 6:38 AM

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KhuongTrang wrote:

Problem. Let $a,b,c$ be non-negative real variables with $a+b+c>0.$ Prove that $\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$ Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$

We have the following estimate
$\frac{12a(a+2b)}{4ab+bc+ca} \geqslant \frac{32a^3+3(33b+56c)a^2+3(26b^2+102bc+13c^2)a-4(4b+c)(b-2c)^2}{11[ab(a+b)+bc(b+c)+ca(c+a)]+51abc}.$

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Zuyong

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Zuyong

#98 h Yesterday at 1:56 PM

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Nguyenhuyen_AG wrote:

KhuongTrang wrote:

Problem. Let $a,b,c$ be non-negative real variables with $a+b+c>0.$ Prove that $\color{black}{\frac{a^2+2ab}{4ab+bc+ca}+\frac{b^2+2bc}{4bc+ca+ab}+\frac{c^2+2ca}{4ca+ab+bc}\ge \frac{3}{2}. }$ Equality holds iff $(a,b,c)\sim(t,t,t)$ or $(a,b,c)\sim\left(t,0,2t\right)$ where $t>0.$

We have the following estimate
$\frac{12a(a+2b)}{4ab+bc+ca} \geqslant \frac{32a^3+3(33b+56c)a^2+3(26b^2+102bc+13c^2)a-4(4b+c)(b-2c)^2}{11[ab(a+b)+bc(b+c)+ca(c+a)]+51abc}.$

How do you come up with this estimate?

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KhuongTrang

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#99 h Today at 7:30 AM

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Problem. Let $a,b,c$ be real variables with $a+b+c=3.$ Prove that $\color{black}{\frac{1}{a^{2}+b^{2}+c}+\frac{1}{b^{2}+c^{2}+a}+\frac{1}{c^{2}+a^{2}+b}\le \frac{15}{2a^{2}+2b^{2}+2c^{2}+9}.}$ Equality holds iff $a=b=c=1$ or $a=b=0,c=3$ and any cyclic permutations.

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Zuyong

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Zuyong

#100 h Today at 12:07 PM

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Why do the author just post inequalities without any solutions?

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