Of course it is wrong. The problem is: Given $a, b, c$ three positive real numbers, prove the following inequality \[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right). \] I have proposed this problem. Remark. This inequality given in the contest is stronger than the following one given at Iran Olympiad in 2005. Given $a, b, c$ three positive real numbers, prove the following inequality \[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \]

yes.but still easy.after computing the ineq is 2sigma(a^2/b^2)+sigma(a/c)>=3sigma(a/b).this is equivalent with 2sigma(a^4c^2)+sigma(a^3b^2c)>=3sigma(a^3bc^2).but 2a^4c^2+ab^3c^2>=3a^3bc^2 from am-gm.writing the other 2 ineq like this one and suming we get the desired ineq

You could also expand and simplify, and the ineq is killed by Muirhead.

Valentin Vornicu wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ \left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 32 \cdot \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) . \] This is very easy! Just use $x^2+y^2+z^2\geq xy+yz+zx\ (x>0,\ y>0,\ z>0)$ and $xyz=1.$ :

$x=\frac{a}{b}$;$y=\frac{b}{c}$;$z=\frac{c}{a}$ $xyz=1$ The inequality becomes: $2(x+y+z)^2 \geq 3(xy+yz+zx) + 3(x+y+z)$ We know $(x+y+z)^2 \geq 3(xy+yz+zx)$ (1) Also by AM-GM: $x+y+z \geq 3$ $(x+y+z)^2 \geq 3(x+y+z)$ (2) By summing (1) and (2) we get $2(x+y+z)^2 \geq 3(xy+yz+zx) + 3(x+y+z)$ which is what we wanted to prove

I used the same notation as M4RIO and got to this: $2(x^2+y^2+z^2)+xy+xz+yz \geq 3(x+y+z)$ which is true considering that $2x^2+yz \geq 3x$

With substitution x=a/b y=b/c z=c/a 2(x+y+z)(x+y+z)>=3(xy+xz+yz)+3(x+y+z)2(x*x+y*y+z*z)+(xy+yz+xz)>=3(x+y+z) X*x+x*x+yz>=3*∛(x*x*x*x*y*z)=∛(x*x*x )=x (y*z=b/a x*x*x*x=a/b*b/a*a/b*a/b*a/b=x*x*x) 2(x*x+y*y+z*z)+(xy+yz+xz)>=3(x+y+z)

Valentin Vornicu wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ ( \frac ab + \frac bc + \frac ca )^2 \geq \frac 32 \cdot ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b ) . \] Note that the left hand side can be rewritten as $\frac{3}{2}\cdot( \frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b})$. Now, the right can be expressed as $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2( \frac{a}{c}+\frac{b}{a}+\frac{c}{b})$. We move the common terms of $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ to the left, giving $\sum_{cyc} \frac{a^2}{b^2}\ge \sum_{cyc} \frac{3}{2}\cdot \frac{a}{b}-\frac{1}{2}\cdot \frac{b}{a}$. We can prove the inequalities by showing that the corresponding terms also satisfy the inequality; i.e. $\frac{a^2}{b^2}\ge \frac{3}{2}\cdot \frac{a}{b}-\frac{1}{2}\cdot \frac{b}{a}$. Setting $\frac{a}{b}=x$, we have $x^2\ge\frac{3}{2}x-\frac{1}{2}\cdot\frac{1}{2}$, which is equivalent to $2x^3-3x^2+x=x(2x^2-3x+1)\ge 0$. Since the discriminant of $F(x)=2x^2-3x+1$, $\Delta=9-8=1$, $F$ will always be positive, so the result follows.

PlatinumFalcon wrote: Valentin Vornicu wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ ( \frac ab + \frac bc + \frac ca )^2 \geq \frac 32 \cdot ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b ) . \] Note that the left hand side can be rewritten as $\frac{3}{2}\cdot( \frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b})$. Now, the right can be expressed as $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2( \frac{a}{c}+\frac{b}{a}+\frac{c}{b})$. We move the common terms of $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ to the left, giving $\sum_{cyc} \frac{a^2}{b^2}\ge \sum_{cyc} \frac{3}{2}\cdot \frac{a}{b}-\frac{1}{2}\cdot \frac{b}{a}$. We can prove the inequalities by showing that the corresponding terms also satisfy the inequality; i.e. $\frac{a^2}{b^2}\ge \frac{3}{2}\cdot \frac{a}{b}-\frac{1}{2}\cdot \frac{b}{a}$. Setting $\frac{a}{b}=x$, we have $x^2\ge\frac{3}{2}x-\frac{1}{2}\cdot\frac{1}{2}$, which is equivalent to $2x^3-3x^2+x=x(2x^2-3x+1)\ge 0$. Since the discriminant of $F(x)=2x^2-3x+1$, $\Delta=9-8=1$, $F$ will always be positive, so the result follows. The ineq. becomes $ x^2\ge\frac{3}{2}x-\frac{1}{2x} $ The generating cubic function should be $f(x)=2x^3-3x^2+1=(x-1)^2(2x+1)\geq 0$ Or AM-GM $2x^3+1=x^3+x^3+1\geq 3\sqrt[3]{x^3.x^3.1}=3x^2$, the result follows.

Letting $x = a/b, y = b/c, z = c/a$, we see we have $xyz = 1$ and $x,y,z$ are positive. Our inequality now becomes \[(x+y+z)^2 \ge \frac{3}{2}(x+y+z+xy+yz+zx)\] Rearranging yields \[2x^2 + 2y^2+2z^2 + xy + yz + zx \ge 3(x+y+z)\] Now we can just AM-GM: we have $x^2 + x^2 + yz \ge 3x$, and cyclically summing, we get the desired inequality.

Valentin Vornicu wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds \[\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 32 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) . \] Let $a,b,c$ be positive real numbers. Prove that\[{\left(\frac{b}{a}+ \frac{a}{b}\right)\left(\frac{c}{b}+ \frac{b}{c}\right)\left(\frac{a}{c}+ \frac{c}{a} \right)}\ge \frac{4}{3}\left(\frac{a+b} {c}+\frac{b+c}{a}+ \frac{c+a}{b}\right)\]Prove that for all positive real numbers $a,b,c$ the following inequality holds $$ \left(a+ \frac 1a\right)\left(b+ \frac 1b \right)\left(c+ \frac 1c \right) \geq \frac 43 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) . $$ Jichen p/5916256821

Prove that for all positive real numbers $a,b,c$ the following inequality holds $$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 95 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b -1\right) $$$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 94 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b -2\right) $$$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac {18}{11} \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b-\frac{1}{2} \right) $$$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq 2 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b - \frac {3}{2}\right) $$$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac{12}{7} \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b - \frac {3}{4}\right) $$$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac {12(11-4\sqrt 6)}{5} \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b-3 \right) $$

sqing wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds $$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 95 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b -1\right) $$

Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac{3}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$Where $k>0.$ $$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac{a+2b}c + \frac{b+2c}a + \frac{c+2a} b$$

Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^3 \geq \frac{9}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$Where $k>0.$ $$\left( \frac ab + \frac bc + \frac ca \right)^3 \geq \frac{a+8b}c + \frac{b+8c}a + \frac{c+8a} b$$

sqing wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac{3}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$Where $k>0.$

sqing wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^3 \geq \frac{9}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$Where $k>0.$

Valentin Vornicu wrote: Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ \left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 32 \cdot \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) . \] h S624 Let $a,b,c>0$. Prove that$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \sqrt{\frac{a+b}{2c}}+ \sqrt{\frac{b+c}{2a}}+ \sqrt{\frac{c+a}{2b}}$$