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Happy Chinese New Year (2)

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sqing

#1 h Feb 12, 2021, 1:03 PM • 1 Y

Y by xyzz

Let $a,b,c$ are positive reals such that $a\ge b+c.$ Prove that
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$ $\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right) \geq \frac{32}{27}$ $\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$ $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}\geq 1$ here

This post has been edited 1 time. Last edited by sqing, Feb 12, 2021, 1:08 PM

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#2 h Feb 14, 2021, 7:00 AM • 1 Y

Y by MrOreoJuice

The first inequality
Proof:
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$ $\Leftrightarrow \sum_{cyc}\frac{a^2}{(b+c)(a+b+c)}\geq\frac{2}{3}$ $\Leftrightarrow \sum_{cyc}[\frac{a}{b+c}-\frac{a}{a+b+c}]\geq\frac{2}{3}$ $\Leftrightarrow \sum_{cyc}\frac{a}{b+c}\geq\frac{5}{3}$ $\Leftrightarrow 1+\frac{a-b-c}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{5}{3}$ $\Leftrightarrow \frac{a-b-c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\geq\frac{8}{3}$ $\Leftarrow \frac{a-b-c}{b+c}+\frac{4(a+b+c)}{2a+b+c}\geq\frac{8}{3}\hspace{3em} (\text{AM-HM inequality})$ $\Leftrightarrow \frac{a-b-c}{b+c}+\frac{12(a+b+c)-8(2a+b+c)}{3(2a+b+c)} \geq 0$ $\Leftrightarrow \frac{a-b-c}{b+c}-\frac{4a-4b-4c}{3(2a+b+c)}\geq 0$ $\Leftrightarrow (a-b-c)[\frac{1}{b+c}-\frac{4}{3(2a+b+c)}]\geq 0$ $\Leftrightarrow (a-b-c)\cdot\frac{6a-b-c}{3(b+c)(2a+b+c)}\geq 0$ $\Leftrightarrow (a-b-c)\cdot[\frac{5a}{3(b+c)(2a+b+c)}+\frac{a-b-c}{3(b+c)(2a+b+c)}]\geq 0$ Equality occurs when $a=2b=2c$ .
Q.E.D.

This post has been edited 1 time. Last edited by Man0308Fong, Feb 15, 2021, 2:37 AM

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sqing

#3 h Feb 14, 2021, 7:34 AM • 1 Y

Y by Mathskidd

Thanks.

sqing wrote:

Let $a,b,c$ are positive reals such that $a\ge b+c.$ Prove that
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$

Solution:
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$ $\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c \geq \frac{5}{3}(a+b+c)$ $\Leftrightarrow \frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b} \geq \frac{5}{3}(a+b+c)$ $\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{5}{3}$ Let $t=\frac{a}{b+c}$ $(t\ge 1)$
$\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{{(b+c)}^2}{a(b+c)+2bc}\ge \frac{{(b+c)}^2}{a(b+c)+\frac{{(b+c)}^2}{2}}=\frac{2}{2t+1}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq t+\frac{2}{2t+1}$ It remains to prove
$t+\frac{2}{2t+1}\ge \frac{5}{3} \iff \frac{(6t-1)(t-1)}{3}\ge 0$ which is true.
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$ Equality occurs when $a=2b=2c$ .

This post has been edited 2 times. Last edited by sqing, Feb 28, 2021, 6:26 AM

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#4 h Feb 14, 2021, 2:18 PM • 3 Y

Y by xyzz, MrOreoJuice, Mathskidd

The second inequality
Proof: Without loss of generality, let $a+b+c=1$ , so $a\geq b+c \Leftrightarrow 1\geq a\geq\frac{1}{2}\geq b,c$
$\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right)\geq\frac{32}{27}$ $\Leftrightarrow \prod_{cyc}\left(\frac{a}{1-a}+\frac{b}{1-b}\right)\geq\frac{32}{27}$ We use Jensen's inequality
$\frac{b}{1-b}+\frac{c}{1-c}=\frac{1}{1-b}+\frac{1}{1-c}-2\geq\frac{2}{\frac{(1-b)+(1-c)}{2}}-2=\frac{2(1-a)}{1+a}$ Meanwhile, we use AM-GM inequality and Jensen's inequality again,
$\left(\frac{a}{1-a}+\frac{b}{1-b}\right)\left(\frac{a}{1-a}+\frac{c}{1-c}\right)=\left(\frac{2a-1}{1-a}+\frac{1}{1-b}\right)\left(\frac{2a-1}{1-a}+\frac{1}{1-c}\right)$ $=\left(\frac{2a-1}{1-a}\right)^2+\left(\frac{2a-1}{1-a}\right)\left(\frac{1}{1-b}+\frac{1}{1-c}\right)+\frac{1}{(1-b)(1-c)}$ $\geq\left(\frac{2a-1}{1-a}\right)^2+\frac{2a-1}{1-a}\cdot\frac{4}{1+a}+\frac{4}{(1+a)^2}$ $=\left(\frac{2a-1}{1-a}+\frac{2}{1+a}\right)^2$ Therefore, we got
$\prod_{cyc}\left(\frac{a}{1-a}+\frac{b}{1-b}\right)\geq\left(\frac{2a-1}{1-a}+\frac{2}{1+a}\right)^2\cdot\frac{2(1-a)}{1+a}$ Denote $f(x)=\left(\frac{2x-1}{1-x}+\frac{2}{1+x}\right)^2\cdot\frac{2(1-x)}{1+x}$
$f'(x)=2\left(\frac{2x-1}{1-x}+\frac{2}{1+x}\right)\cdot\frac{1}{(1+x)^2}\cdot\frac{x(5-3x)}{(1-x)(1+x)}$ When $x\in[\frac{1}{2},1]$ , $\left(\frac{2x-1}{1-x}+\frac{2}{1+x}\right)>0$ , $\frac{1}{(1+x)^2}>0$ , $\frac{x(5-3x)}{(1-x)(1+x)}>0$
Therefore $f'(x)>0$ .
In other word, $f(a) \geq f(\frac{1}{2}) = \left(\frac{4}{3}\right)^2\times\frac{2}{3}=\frac{32}{27}$
$\left(\frac{a}{b+c}+\frac{b}{c+a}\right)\left(\frac{b}{c+a}+\frac{c}{a+b}\right)\left(\frac{c}{a+b}+\frac{a}{b+c}\right)\geq f(a) \geq f(\frac{1}{2})=\frac{32}{27}$ Equality holds when $a=2b=2c$ .
Q.E.D.

This post has been edited 1 time. Last edited by Man0308Fong, Feb 15, 2021, 2:38 AM

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#5 h Feb 14, 2021, 3:09 PM • 2 Y

Y by xyzz, Mango247

The third inequality
Proof: Denote $k=a-b-c \geq 0$
$\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$ $\Leftrightarrow 9\sum_{cyc}bc(b+c)-7(a+b)(b+c)(c+a)\geq 0$ $\Leftrightarrow 9[bc(b+c)+c(b+c+k)(b+2c+k)+(b+c+k)b(2b+c+k)]-7(b+c)(b+2c+k)(2b+c+k)\geq 0$ $\Leftrightarrow 4b^3+4c^3-4b^2c-4bc^2+2k^2(b+c)+6k(b^2-bc+c^2)\geq 0$ $\Leftrightarrow 2(b+c)(b-c)^2+k^2(b+c)+3k(b^2-bc+c^2)\geq 0$ Equality holds when $k=0$ and $b=c$ , i.e. $a=2b=2c$ .
Q.E.D

This post has been edited 1 time. Last edited by Man0308Fong, Feb 14, 2021, 3:19 PM

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sqing

#6 h Feb 15, 2021, 1:01 AM

Y by

Thanks.
Let $a, b$ and $c$ be non-negative numbers such that $a \geq b+c.$ Prove that
$\frac{a}{b+c} + \frac{b}{c+a} +\frac{2c}{a+b} \geq \frac{4\sqrt 2}{3}$

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azt2

#7 h Feb 15, 2021, 1:06 AM • 1 Y

Y by Yeetopia

It's technically not "Chinese" New Year. The correct term is "Lunar" New Year. A lot of countries follow the lunar calendar, not just China. So TECHNICALLY it's not Chinese New Year.

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sqing

#8 h Feb 15, 2021, 1:10 AM

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Let $a, b$ and $c$ be non-negative numbers such that $a \geq b+c.$ Prove that
$\frac{a}{b+c} + \frac{b}{c+a} +\frac{2c}{a+b} \geq \frac{4\sqrt 2}{3}$ Equality holds when $a=\frac{4+3\sqrt 2}{2}c, b=\frac{2+3\sqrt 2}{2}c.$
Let $a,b,c>0$ and $a\geq b+c.$ Prove that
$\frac{a}{b+c} \left( \frac{b}{c+a}+\frac{c}{a+b} \right) \geq \frac{b}{2c+b} + \frac{c}{c+2b}\geq\frac{2}{3}$ $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq 1+\frac{b}{2c+b} + \frac{c}{c+2b}$ h
Thanks. Just a title or symbol

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#9 h Feb 15, 2021, 2:35 AM

Y by

The fourth inequality
Proof:
$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}\geq 1$ $\Leftrightarrow \sum_{cyc}\sqrt{bc(b+c)}\geq\sqrt{(a+b)(b+c)(c+a)}$ $\Leftrightarrow \sum_{cyc}bc(b+c)+2\sqrt{abc}\sqrt{c(b+c)(c+a)}\geq(a+b)(b+c)(c+a)=\sum_{sym}a^2b+2abc$ $\Leftrightarrow \sum_{sym}b^2c+2\sqrt{abc}\sqrt{c(b+c)(c+a)}\geq(a+b)(b+c)(c+a)=\sum_{sym}a^2b+2abc$ $\Leftrightarrow \sum_{cyc}\sqrt{c(b+c)(c+a)}\geq\sqrt{abc}$ In the meantime,
$\sum_{cyc}\sqrt{c(b+c)(c+a)}\geq\sum_{cyc}{c\cdot 2\sqrt{bc} \cdot 2\sqrt{ca}}=2\sum_{cyc}c\sqrt[4]{ab}\geq 6\sqrt[3]{c\sqrt[4]{ab}\cdot a\sqrt[4]{bc}\cdot b\sqrt[4]{ca}}=6\sqrt{abc}>\sqrt{abc}$ Equality holds when $abc=0$ , i.e. equality can't occur.
Q.E.D.

##The inequality should be $\forall a,b,c \in \mathbb{R}^{+}$ , we got
$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}>1$

This post has been edited 2 times. Last edited by Man0308Fong, Feb 15, 2021, 3:08 AM

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sqing

#10 h Feb 15, 2021, 2:46 AM

Y by

Thanks.
Show that for $a,b,c > 0$ the following inequality holds:
$\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$ Let $a,b,c$ be positive real numbers. Prove that $\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b}\leq \frac{a+b+c}{4}$ Let $a$ , $b$ , $c$ be positive reals such that $a+b+c=3$ . Show that $\frac{ab}{2a+b+c} + \frac{bc}{2b+c+a}+ \frac{ca}{2c+a+b}\leq \frac{3}{4}$ Very easy.
$\implies$ $\sqrt{\frac{ab}{2a+b+c}} + \sqrt{\frac{bc}{2b+c+a}} + \sqrt{\frac{ca}{2c+a+b}} \leq \frac{3}{2}$ (Crux (3)2021,4624)
Let $a$ , $b$ , $c$ be positive reals such that $a+b+c=3$ . Show that $\frac{ab}{ka+b+c} + \frac{bc}{kb+c+a}+ \frac{ca}{kc+a+b}\leq \frac{3}{k+2}$ Where $k\geq1 .$
$\implies$ Let $a$ , $b$ , $c$ be positive reals such that $a+b+c=3$ . Show that
$\sqrt{\frac{ab}{ka+b+c}} + \sqrt{\frac{bc}{kb+c+a}} + \sqrt{\frac{ca}{kc+a+b}} \leq \frac{3}{\sqrt{k+2}}$ Where $k\geq1 .$
Solution of Zhangyanzong:

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#12 h Mar 22, 2021, 11:53 AM

Y by

sqing wrote:

Thanks.

sqing wrote:

Let $a,b,c$ are positive reals such that $a\ge b+c.$ Prove that
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$

Solution:
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$ $\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c \geq \frac{5}{3}(a+b+c)$ $\Leftrightarrow \frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b} \geq \frac{5}{3}(a+b+c)$ $\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{5}{3}$ Let $t=\frac{a}{b+c}$ $(t\ge 1)$
$\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{{(b+c)}^2}{a(b+c)+2bc}\ge \frac{{(b+c)}^2}{a(b+c)+\frac{{(b+c)}^2}{2}}=\frac{2}{2t+1}$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq t+\frac{2}{2t+1}$ It remains to prove
$t+\frac{2}{2t+1}\ge \frac{5}{3} \iff \frac{(6t-1)(t-1)}{3}\ge 0$ which is true.
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{2}{3}(a+b+c)$ Equality occurs when $a=2b=2c$ .

Dear Sir Sqing ,
When did you complete the proof?

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sqing

#13 h Mar 22, 2021, 1:34 PM

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Let $a,b,c$ are positive reals such that $a\ge b+c.$ Prove that
$\left ( a+b+c\right )\left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )\geq 10$ $\left ( a^2+b^2+c^2\right )\left (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right )\geq \frac{27}{2}$ $\left ( a^3+b^3+c^3\right )\left (\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3} \right )\geq \frac{85}{4}$ $\left ( a^4+b^4+c^4\right )\left (\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} \right )\geq \frac{297}{8}$

sqing wrote:

Let $a,b,c$ are positive reals such that $a\ge b+c.$ Prove that
$1>\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ ab}{(b+c)(c+a)} \geq \frac{7}{9}$

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sqing

#14 h Mar 28, 2021, 2:29 AM

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Let $a,b,c$ are non-negative numbers such that $a\ge 2(b+c).$ Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 2+\frac{a^2+b^2+c^2}{5(ab+bc+ca)}$ here

sqing wrote:

Let $a,b,c$ are positive reals such that $a\ge b+c.$ Prove that
$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}}\geq 1$

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Let $a,b,c$ be positive numbers. Probe that
$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ ab}{(b+c)(c+a)}} \geq \frac{9 a b c(a+b+c)}{2(a b+b c+c a)^{2}}$

This post has been edited 3 times. Last edited by sqing, Aug 8, 2022, 2:04 PM

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