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 $k < 0$.

$LHS-6=(\frac{x}{y}+\frac{y}{x}-2)+(\frac{x}{z}+\frac{z}{x}-2)+(\frac{y}{z}+\frac{z}{y}-2)=$ $=\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 $ $\ge \frac{((x-y)+(x-z))^2}{xy+xz}\ge \frac{x^2}{x(y+z)}\ge 1$

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 $x\geq y + z$

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 \[\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geqslant 7.\]

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 $x:y:z=2:1:1.$

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

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 $a_1,a_2,\cdots,a_n (n\ge 3)$ 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 $S=a_1+a_2+a_3+\cdots+a_n.$

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 $x:y:z=2:1:1.$

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 $x=2y=2z.$

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 $\frac{x}{b} \ge x$.

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)

Let $a, b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that

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 $$\frac{a}{b+c} + \frac{b}{c+a} +\frac{2c}{a+b} \geq \frac{4\sqrt 2}{3} $$

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:

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

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)

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)

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

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

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

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 $x:y:z=2:1:1.$

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

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

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

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

sqing wrote: Let $a,b,c$ be positive real numbers such that $ a \geq 2(b+c)$. Prove that $$\frac{b}{a+b}+\frac{c}{c+a} \leq \frac{2}{5}$$ $\frac b{a+b}+\frac c{c+a}\leq\frac b{3b+2c}+\frac c{2b+3c}=\frac{2b^2+6bc+2c^2}{6b^2+13bc+6c^2}\leq\frac25$

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 $a\geq2x$.

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

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 $x=y+z=2y$.

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 $$(a^2+2bc)\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right) \geq \frac{49}{9}$$

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)

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

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

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