Let $ a,b,c$ be positive real numbers such that $8(a^2 + b^2 + c^2) =11(ab + bc + ca).$ Prove that $$ \frac {54}{5}\leq (a + b + c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \leq \frac {27}{2}$$$$\frac {69}{40}\leq \frac{a}{b+c} + \frac{b}{c + a} +\frac{c}{a + b} \leq \frac {51}{28}$$$$\frac {39}{5}\leq \frac{b+c}{a} + \frac{c + a}{b} +\frac{a + b} {c}\leq \frac {21}{2}$$$$ \frac {141}{20}\leq \frac{a^2}{bc} + \frac{b^2}{ca} +\frac{c^2}{ab}\leq \frac {129}{16}$$

Let $a, b, c$ be positive real numbers. Prove that$$8\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right)\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) \le 9 \left(1+\frac{a}{b} \right)\left(1+\frac{b}{c} \right)\left(1+\frac{c}{a} \right)$$

Let $a, b, c$ be positive real numbers. Prove that $$\left( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\right) \left( \frac{b}{a}+ \frac{c}{b}+ \frac{a}{c}\right)\geq\left(\frac{a}{b}+\sqrt[3]{9}-1\right) \left(\frac{b}{c}+\sqrt[3]{9}-1\right)\left(\frac{c}{a}+\sqrt[3]{9}-1\right)$$