$\frac{a+b}{c^2}+ \frac{c+a}{b^2}+ \frac{b+c}{a^2} = (a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq \frac{a+b+c}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq \frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Let $a, b$ and $c$ be positive real numbers. Show that $$ \frac{b+c}{a^2}+\frac{c+a}{b^2}+ \frac{a+b}{c^2}+\ge \frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 4\left(\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}\right)$$

parmenides51 wrote: \[\frac{a+b}{c^2}+ \frac{c+a}{a^2}+ \frac{b+c}{a^2}\ge \frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\] By AM-GM, \[\frac{a}{b^2} + \frac{b}{a^2} = \left(\frac{1}{3} \cdot \frac{a}{b^2} + \frac{2}{3} \cdot \frac{b}{a^2}\right) + \left(\frac{2}{3} \cdot \frac{a}{b^2} + \frac{1}{3} \cdot \frac{b}{a^2}\right) \ge \frac{1}{a} + \frac{1}{b}\]By summing similar inequalities and by AM-HM we get \[\frac{a+b}{c^2}+ \frac{c+a}{a^2}+ \frac{b+c}{a^2}\ge 2 \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \ge \frac{9}{a+b+c} + \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\]We see that equality holds if and only if $a=b=c$. sqing wrote: \[\frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge 4\left(\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}\right)\] Multiplying both sides by $a+b+c$ then simplifying, we see this is equivalent to \[\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \ge 4 \left(\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}\right)\]which follows from AM-HM by summing the inequalities of the form \[\frac{c}{a} + \frac{c}{b} \ge \frac{4c}{a+b}\]

Rioplatense Olympiad 2002 level 2 P1: Let $a, b$ and $c$ be positive real numbers. Prove that $$ \big(\frac{a}{b+c}+\frac{1}{2}\big) \big(\frac{b}{c+a}+\frac{1}{2}\big) \big(\frac{c}{a+b}+\frac{1}{2}\big)\ge 1$$Rioplatense Mathematical Olympiad 2002 Let $a, b$ and $c$ be positive real numbers. Prove that $$ \big(\frac{a}{a+b+c}+\frac{2}{3}\big) \big(\frac{b}{a+b+c}+\frac{2}{3}\big) \big(\frac{c}{a+b+c}+\frac{2}{3}\big)\le 1$$

Let $a, b$ and $c$ be positive real numbers. Show that $$\frac{a+b}{c^2}+ \frac{c+a}{a^2}+ \frac{b+c}{a^2}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge \frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$h Let $a, b$ and $c$ be positive real numbers such that $a+b+c=3.$ Show that$$\frac{a+b}{c^3}+ \frac{c+a}{a^3}+ \frac{b+c}{a^3}\ge2(a^2+b^2+c^2)$$

parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers. Show that $\frac{a+b}{c^2}+ \frac{c+a}{a^2}+ \frac{b+c}{a^2}\ge \frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ I think there is a typo, it should be $\frac{c+a}{b^2}$ instead of $\frac{c+a}{a^2}$.

jasperE3 wrote: parmenides51 wrote: Let $a, b$ and $c$ be positive real numbers. Show that $\frac{a+b}{c^2}+ \frac{c+a}{a^2}+ \frac{b+c}{a^2}\ge \frac{9}{a+b+c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ I think there is a typo, it should be $\frac{c+a}{b^2}$ instead of $\frac{c+a}{a^2}$. Indeed, I had that typo as I checked here

Rioplatense Olympiad 2002 level 2 P1 Let $a,b,c>0.$ Prove that$$(\frac{a}{b+c}+\frac{1}{2})(\frac{b}{c+a}+\frac{1}{2})(\frac{c}{a+b}+\frac{1}{2})\geq 1$$