Problem 1 - Inequality

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Problem 1 - Inequality

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Source: 2020 Austrian National Competition for Advanced Students, Part 1 problem 1

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Ln142

#1 h Jun 6, 2020, 1:45 PM • 1 Y

Y by rightways

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ .
Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur?

(Walther Janous)

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#2 h Jun 6, 2020, 2:02 PM

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Let $x = y + z + k$ where $k \ge 0$ . Then
$LHS = 2 + \frac{2y + k}{z} + \frac{y + z}{y + z + k} + \frac{2z + k}{y} \ge 7 + \frac{k}{z} + \frac{k}{y} - \frac{k}{y+z+k} \ge 7 + \frac{4k}{y + z} - \frac{k}{y + z + k} = 7 + k \left( \frac{4}{y + z} - \frac{1}{y + z + k} \right) \ge 7$ Equality occur when $k = 0$ , or $\frac{4}{y + z} - \frac{1}{y + z + k} = 0$ , but the latter gives $k < 0$ .

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#3 h Jun 6, 2020, 2:06 PM • 2 Y

Y by Illuzion, Mango247

Hint
Solution

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mudok

#4 h Jun 6, 2020, 8:37 PM • 5 Y

Y by Math00954, Illuzion, star-1ord, Math-wiz, KhaliLCuy

$LHS-6=(\frac{x}{y}+\frac{y}{x}-2)+(\frac{x}{z}+\frac{z}{x}-2)+(\frac{y}{z}+\frac{z}{y}-2)=$

$=\frac{(x-y)^2}{xy}+\frac{(x-z)^2}{xz}+\frac{(y-z)^2}{yz}\ge \frac{(x-y)^2}{xy}+\frac{(x-z)^2}{xz}\ge$

$\ge \frac{((x-y)+(x-z))^2}{xy+xz}\ge \frac{x^2}{x(y+z)}\ge 1$

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SBM

#5 h Jun 7, 2020, 12:48 AM

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Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ .
Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur?

(Walther Janous)

We have $:$ $\text{LHS}-\text{RHS}={\frac { \left( y+z \right) \left( x-y-z \right) ^{2}}{zxy}}+{\frac { \left( {y}^{2}+yz+{z}^{2} \right) \left( x-y-z \right) }{zxy}}+\,{ \frac { 2\left( y-z \right) ^{2}}{yz}} \geq 0$ Which is clearly true for $x\geq y + z$

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#7 h Jun 7, 2020, 2:35 AM

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Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ .
Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur?
(Walther Janous)

We have
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} - 7 =\frac{(x-y-z)[x(y+z)-yz]}{xyz}+\frac{2(y-z)^2}{yz},$ and $x(y+z) \geqslant (y+z)^2 > yz.$ Thefore
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geqslant 7.$

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sqing

#8 h Jun 7, 2020, 8:24 AM • 2 Y

Y by Mathsolver19, Wizard0001

Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ . Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur? (Walther Janous)

WLOG $x+y+z=1,$ so $1>x\ge \frac{1}{2},$ $(2x-1)(5x-1)\ge 0\iff\frac{1}{x}+\frac{4}{y+z}\ge 10$ $\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} =\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3\geq \frac{1}{x}+\frac{4}{y+z}-3\geq 7.$ Equality holds when $x:y:z=2:1:1.$

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#9 h Jun 7, 2020, 9:17 AM

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sqing wrote:

Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ . Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur? (Walther Janous)

WLOG $x+y+z=1,$ so $1>x\ge \frac{1}{2},$ $(2x-1)(5x-1)\ge 0\iff\frac{1}{x}+\frac{4}{y+z}\ge 10$ $\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} =\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-3\geq \frac{1}{x}+\frac{4}{y+z}-3\geq 7.$ Equality holds when $x:y:z=2:1:1.$

Yay..
Squing can you post some new problems

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#10 h Jun 7, 2020, 9:19 AM

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Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Find the minimum value of $\frac{a+b+c}{d} + \frac{b+c+d}{a} +\frac{c+d+a}{b}+\frac{d+a+b}{c} .$

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sqing

#11 h Jun 7, 2020, 9:33 AM

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Let $a_1,a_2,\cdots,a_n (n\ge 3)$ bepositive real numbers such that $a_1\geq a_2+a_3+\cdots+a_n.$ Prove or disprove $\frac{S-a_1}{a_1} +\frac{S-a_2}{a_2} +\cdots+\frac{S-a_n}{a_n} \geq2n^2-5n+4.$ Where $S=a_1+a_2+a_3+\cdots+a_n.$

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#12 h Jun 8, 2020, 12:40 AM

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Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ . Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur? (Walther Janous)

$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} =x\left(\frac{1}{y}+\frac{1}{z}\right)+ \frac{y+z}{x} +\frac{z}{y}+\frac{y}{z}$ $\geq\frac{4x}{y+z}+ \frac{y+z}{x} +\frac{z}{y}+\frac{y}{z} \geq 7\sqrt[7]{\left(\frac{x}{y+z}\right)^4\cdot \frac{y+z}{x} \cdot \frac{z}{y}\cdot \frac{y}{z} } \geq 7.$ Equality holds when $x:y:z=2:1:1.$

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#13 h Jun 10, 2020, 6:44 AM

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Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ .
Proof that $\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur? (Walther Janous)

Let $x=k(y+z)$ $(k\geq 1.)$ Hence
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} =\frac{1}{k} +2k+(k+1)(\frac{y}{z} +\frac{z}{y} ) \geq \frac{1}{k} +4k+2\geq 7.$ Equality holds when $x=2y=2z.$

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#14 h Jun 14, 2020, 2:49 AM

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sqing wrote:

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Find the minimum value of $\frac{a+b+c}{d} + \frac{b+c+d}{a} +\frac{c+d+a}{b}+\frac{d+a+b}{c} .$

Solution of wdym:
WLOG let $a=1=b+c+d+x$ . Then,
$\begin{align*} \frac{a+b+c}{d}+\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c} &= \frac{2b+2c+d+x}{d}+b+c+d+\frac{b+2c+2d+x}{b}+\frac{2b+c+2d+x}{c} \\ &=3+\frac{2b}{c}+\frac{2c}{b}+\frac{2b}{d}+\frac{2d}{b}+\frac{2c}{d}+\frac{2d}{c}+1-x+\frac{x}{b}+\frac{x}{c}+\frac{x}{d} \\ &\ge 3+12+1-x+\frac{x}{b}+\frac{x}{c}+\frac{x}{d} \\ &\ge 16, \\ \end{align*}$ by AM-GM and that $\frac{x}{b} \ge x$ .

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#15 h Jun 15, 2020, 2:18 AM

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Let $a, b$ and $c$ be positive real numbers such that $a \geq b+c.$ Proof that
$\frac{2a}{3(b+c)} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{4}{3} .$
Let $a_1,a_2,\cdots,a_n (n\ge 3)$ bepositive real numbers such that $a_1\geq a_2+a_3+\cdots+a_n.$ Prove that $\frac{a_1} {S-a_1}+\frac{a_2}{S-a_2}+\cdots+\frac{a_n}{S-a_n}\geq \frac{3n-4}{2n-3} .$ Where $S=a_1+a_2+a_3+\cdots+a_n.$
(Lijvzhi)

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#16 h Jun 15, 2020, 2:33 AM

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Let $a, b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that
$\frac{3a}{4(b+c+d)} + \frac{b}{c+d+a} +\frac{c}{d+a+b} +\frac{d}{a+b+c} \geq \frac{3}{2} .$ n

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#17 h Jun 15, 2020, 2:48 AM

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sqing wrote:

Let $a, b$ and $c$ be positive real numbers such that $a \geq b+c.$ Proof that
$\frac{2a}{3(b+c)} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{4}{3} .$

$\frac{2a}{3(b+c)} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{2a}{3(b+c)} + \frac{(b+c)^2}{2bc+a(b+c)}\geq2\left( \frac{2a+b+c}{9(b+c)} + \frac{b+c}{2a+b+c}\right)\geq\frac{4}{3} .$ Let $a, b$ and $c$ be positive real 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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#18 h Jun 21, 2020, 3:38 PM

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sqing wrote:

Let $a_1,a_2,\cdots,a_n (n\ge 3)$ bepositive real numbers such that $a_1\geq a_2+a_3+\cdots+a_n.$ Prove or disprove $\frac{S-a_1}{a_1} +\frac{S-a_2}{a_2} +\cdots+\frac{S-a_n}{a_n} \geq2n^2-5n+4.$ Where $S=a_1+a_2+a_3+\cdots+a_n.$

Solution of Zhangyanzong:

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#19 h Jun 21, 2020, 4:50 PM • 3 Y

Y by Mango247, Mango247, Mango247

sqing wrote:

sqing wrote:

Let $a_1,a_2,\cdots,a_n (n\ge 3)$ bepositive real numbers such that $a_1\geq a_2+a_3+\cdots+a_n.$ Prove or disprove $\frac{S-a_1}{a_1} +\frac{S-a_2}{a_2} +\cdots+\frac{S-a_n}{a_n} \geq2n^2-5n+4.$ Where $S=a_1+a_2+a_3+\cdots+a_n.$

Solution of Zhangyanzong:

i cannot comprehend the last line

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#20 h Jun 30, 2020, 6:11 AM • 1 Y

Y by Mango247

Let $a, b$ and $c$ be positive real numbers such that $a \geq b+c.$ Proof that
$\frac{a}{b+c} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{5}{3}$ $\frac{a}{b+c} + \frac{b}{c+a} +\frac{4c}{a+b} \geq 2$ $\frac{a}{b+c} + \frac{b}{c+a} +\frac{c}{a+b}+\frac{2(ab+bc+ca)}{a^2+b^2+c^2} \geq 3$ $\frac{b+c}{a} + \frac{c+a}{b}+\frac{a+b}{c}+\frac{6(ab+bc+ca)}{a^2+b^2+c^2} \geq 12$ (Lijvzhi)

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#21 h Jul 7, 2020, 6:34 AM

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Let $a, b$ and $c$ be non-negative numbers such that $a \geq b+c.$ Proof that
$\frac{a}{b+c} + \frac{b}{c+a} +\frac{kc}{a+b} \geq 1+k$ Where $0\leq k< \frac{1}{4}$
$\frac{a}{b+c} + \frac{b}{c+a} +\frac{kc}{a+b} \geq \frac{4\sqrt k-k+2}{3}$ Where $\frac{1}{4} \leq k<4$
$\frac{a}{b+c} + \frac{b}{c+a} +\frac{kc}{a+b} \geq2$ Where $k\ge 4$
(Lijvzhi)

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#22 h Jul 7, 2020, 7:01 AM

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This is basically the same problem as IMO 2004, P4.

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#23 h Jul 7, 2020, 2:44 PM

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Gryphos wrote:

This is basically the same problem as IMO 2004, P4.

Thanks.
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7\iff(x+y+z)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 10$
Since $(a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right)=10$ ,we give: $\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=7$ and $(a+b)(b+c)(c+a)=9abc$
h

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#24 h Nov 3, 2020, 6:54 AM • 3 Y

Y by Mango247, Mango247, Mango247

sqing wrote:

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Find the minimum value of $\frac{a+b+c}{d} + \frac{b+c+d}{a} +\frac{c+d+a}{b}+\frac{d+a+b}{c} .$

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that $\frac{a+b+c}{d} + \frac{b+c+d}{a} +\frac{c+d+a}{b}+\frac{d+a+b}{c}\geq 16 .$ h
Let $a_1,a_2,\cdots,a_n (n\ge 3)$ bepositive real numbers such that $a_1\geq a_2+a_3+\cdots+a_n.$ Prove that $\frac{S-a_1}{a_1} +\frac{S-a_2}{a_2} +\cdots+\frac{S-a_n}{a_n} \geq2n^2-5n+4.$ Where $S=a_1+a_2+a_3+\cdots+a_n.$
Zhanyanzong

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#25 h Nov 22, 2020, 12:45 PM

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sqing wrote:

Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ . Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur? (Walther Janous)

$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} =x\left(\frac{1}{y}+\frac{1}{z}\right)+ \frac{y+z}{x} +\frac{z}{y}+\frac{y}{z}$ $\geq\frac{4x}{y+z}+ \frac{y+z}{x} +\frac{z}{y}+\frac{y}{z} \geq 7\sqrt[7]{\left(\frac{x}{y+z}\right)^4\cdot \frac{y+z}{x} \cdot \frac{z}{y}\cdot \frac{y}{z} } \geq 7.$ Equality holds when $x:y:z=2:1:1.$

$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} =\frac{z}{y}+\frac{y}{z} +x\left(\frac{1}{y}+\frac{1}{z}\right)+ \frac{y+z}{x}$ $\geq 2+\frac{4x}{y+z}+ \frac{y+z}{x} \geq 2+5\sqrt[5]{\left(\frac{x}{y+z}\right)^4\cdot \frac{y+z}{x} } \geq 7.$ Equality holds when $x:y:z=2:1:1.$

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#26 h Feb 12, 2021, 1:10 PM • 2 Y

Y by Mango247, Mango247

Let $a_1,a_2,\cdots,a_n (n\ge 3)$ bepositive real numbers such that $a_1\geq a_2+a_3+\cdots+a_n.$ Prove that $\frac{S-a_1}{a_1} +\frac{S-a_2}{a_2} +\cdots+\frac{S-a_n}{a_n} \geq2n^2-5n+4.$ Where $S=a_1+a_2+a_3+\cdots+a_n.$
https://artofproblemsolving.com/community/c6h2449188p20353975
Crux, Vol. 45(5), May 2019 ; Crux , Vol. 45(10), December 2019

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#27 h Apr 30, 2021, 12:58 AM

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Let $a,b,c$ are positive real numbers such that $a \geq b+c .$ Prove that $(a^n + b^n + c^n)(\frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n})\geq 2^{n+1}+\frac{1}{2^{n-1}}+5.$ Where $n\in N^+.$
Let $a,b,c$ are positive real numbers such that $a \geq b+c .$ Prove that $(a^2 + b^2 + c^2)(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2})\geq \frac{27}{2}.$ here

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#28 h Nov 17, 2021, 1:11 PM

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Let $a,b,c$ be positive real numbers such that $a \geq 2(b+c)$ . Prove that
$\frac{a+b}{c} + \frac{b+c}{a} +\frac{c+a}{b} \geq \frac{21}{2}$ $\frac{a}{b+c} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{12}{5}$ $\frac{2a}{3(b+c)} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{26}{15}$ $\frac{a}{b+c} + \frac{b}{c+a} +\frac{2c}{a+b} \geq \frac{4+6\sqrt 2}{5}$ $(a^2 + b^2 + c^2)(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2})\geq \frac{297}{8}$

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#29 h Nov 17, 2021, 1:54 PM

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Let $a,b,c$ be positive real numbers such that $a \geq 2(b+c)$ . Prove that
$\frac{b}{a+b}+\frac{c}{c+a} \leq \frac{2}{5}$ $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{7}{5}(a+b+c)$

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DottedCaculator

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#31 h Nov 17, 2021, 2:07 PM

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sqing wrote:

Let $a,b,c$ be positive real numbers such that $a \geq 2(b+c)$ . Prove that
$\frac{b}{a+b}+\frac{c}{c+a} \leq \frac{2}{5}$

$\frac b{a+b}+\frac c{c+a}\leq\frac b{3b+2c}+\frac c{2b+3c}=\frac{2b^2+6bc+2c^2}{6b^2+13bc+6c^2}\leq\frac25$

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#32 h Nov 17, 2021, 2:13 PM

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Beautiful. Thanks.

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#33 h Nov 17, 2021, 2:19 PM

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sqing wrote:

Let $a,b,c$ be positive real numbers such that $a \geq 2(b+c)$ . Prove that
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{7}{5}(a+b+c)$

$\frac{(10a)^2}{100b+100c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{(10a+b+c)^2}{2a+101b+101c}$ . Let $x=b+c$ . Then, it suffices to show $\frac{(10a+x)^2}{2a+101x}\geq\frac75(a+x)$ . This is equivalent to $500a^2+100ax+5x^2\geq7(a+x)(2a+101x)=14a^2+721ax+707x^2$ , or $486a^2-621ax-702x^2\geq0$ . This factors as $(a-2x)(486a+351x)\geq0$ , which is true since $a\geq2x$ .

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#34 h Nov 17, 2021, 2:23 PM

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Very nice. Thanks.
Let $a,b,c>0$ and $a\geq b+c .$ Prove that
$\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \geq 5$ $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{1+\sqrt{13+16\sqrt{2}}}{2}$ Let $a,b,c$ be positive real numbers such that $a=3(b+c)$ . Prove that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq \frac{ 965}{266}$ Let $a,b,c$ be positive real numbers such that $a=b+c$ . Prove that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq \frac{5+4\sqrt 6}{6}$ Let $a,b,c$ be positive real numbers such that $a=2(b+c)$ . Prove that
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{ab+bc+ca}{a^2+b^2+c^2}\geq \frac{2(13+3\sqrt{15})}{17}$ Let $a,b,c$ be positive real numbers such that $a \geq b+c$ . Prove that
$\frac{a}{b+c}+\frac{2b}{c+a}+\frac{c}{a+b}\geq \frac{4\sqrt 2}{3}$ $\frac{a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}\geq \frac{7}{3}$ $\frac{a}{b+c}+\frac{2b}{c+a}+\frac{3c}{a+b}\geq \sqrt {2}+\sqrt {3}+\sqrt {6}-3$ $\frac{a}{b+c}+\frac{2b}{c+a}+\frac{4c}{a+b} \geq 3\sqrt 2-\frac{3}{2}$ h h h h h

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DottedCaculator

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DottedCaculator

#35 h Nov 17, 2021, 2:29 PM • 1 Y

Y by MihaiT

Ln142 wrote:

Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$ .
Proof that
$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$ When does equality occur?

(Walther Janous)

We have $\frac{x+y}z+\frac{y+z}x+\frac{z+x}y\geq\frac{(2x+2y)^2}{4xz+4yz}+\frac{(y+z)^2}{xy+xz}+\frac{(2z+2x)^2}{4xy+4yz}\geq\frac{(4x+3y+3z)^2}{5xy+8yz+5zx}$ , so it suffices to show $\frac{(4x+3y+3z)^2}{5xy+8yz+5zx}\geq7$ , or $(4x+3y+3z)^2=16x^2+24x(y+z)+9(y+z)^2\geq35x(y+z)+56yz$ . This is equivalent to $16x^2-11x(y+z)+9(y+z)^2\geq56yz$ . Since $56yz\leq14(y+z)^2$ , we need to show $16x^2-11x(y+z)-5(y+z)^2\geq0$ , or $(x-(y+z))(16x+5(y+z))\geq0$ . This is true since $x\geq y+z$ . Equality occurs when $y=z$ and $x=y+z=2y$ .

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sqing

#36 h Nov 17, 2021, 2:44 PM

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Let $a,b,c$ be positive real numbers such that $a \geq b+c$ . Prove that
$(ab+bc+ca)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right) \geq \frac{85}{36}$ Let $a,b,c$ be nonnegative real numbers such that $a \geq b+c$ . Prove that
$(a^2+2bc)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right) \geq \frac{9}{4}$ Let $a,b,c$ be nonnegative real numbers such that $a \geq 2(b+c)$ . Prove that
$(a^2+2bc)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right) \geq \frac{49}{9}$
Good.
Let $a,b,c$ be positive real numbers such that $a \geq 2(b+c)$ . Prove that
$(ab+bc+ca)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right) \geq \frac{49}{18}$ (Mathematical Reflections 1 (2019) O475)

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sqing

#37 h Nov 17, 2021, 2:47 PM

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Let $a,b,c$ be positive real numbers such that $a \geq b+c$ . Prove that
$\frac{a+b}{c} + \frac{b+c}{2a} +\frac{c+a}{b} \geq \frac{13}{2}$ $\frac{a+b}{c} + \frac{b+c}{2a} +\frac{c+a}{2b} \geq 2(\sqrt 2+1)$ $\frac{a+b}{c} + \frac{b+c}{2a} +\frac{c+a}{3b} \geq \frac{11}{6} +\frac{4}{\sqrt 3}$

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sqing

#38 h Nov 21, 2021, 2:51 AM

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sqing wrote:

Let $a,b,c$ be positive real numbers such that $a \geq 2(b+c)$ . Prove that
$\frac{b}{a+b}+\frac{c}{c+a} \leq \frac{2}{5}$ $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{7}{5}(a+b+c)$

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#39 h Nov 21, 2021, 6:40 AM

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sqing wrote:

Let $a,b,c$ be positive real numbers such that $a \geq b+c$ . Prove that
$\frac{a+b}{c} + \frac{b+c}{2a} +\frac{c+a}{b} \geq \frac{13}{2}$

Solution of Zhang yanzong:
$\frac{a+b}{c} + \frac{b+c}{2a} +\frac{c+a}{b} =\frac{a}{c} +\frac{a}{b}+ \frac{b+c}{2a} +\frac{c}{b}+ \frac{b}{c}$ $\geq\frac{4a} {b+c}+\frac{b+c}{2a} +2 =\frac{a} {2(b+c)}+ \frac{b+c}{2a} +\frac{7a} {2(b+c)}+2\geq 1+\frac{7}{2} +2=\frac{13}{2}$

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