Hödler inequallity!

Let $ a, b, c$ be positive real numbers. Prove that $$\frac{(a^2+2b^2+c^2)^3}{(a^3+b^3+c^3)^2} \le 10$$$$\frac{(a^2+b^2+c^2)^3}{(a^3+2b^3+c^3)^2} \le\frac{9}{4}$$$$\frac{(a^2+2b^2+c^2)^3}{(a^3+2b^3+c^3)^2} \le 4$$$$\frac{(a^2+3b^2+c^2)^3}{(a^3+b^3+c^3)^2} \le 29$$$$\frac{(a^2+b^2+c^2)^3}{(a^3+3b^3+c^3)^2} \le\frac{19}{9}$$$$\frac{(a^2+3b^2+c^2)^3}{(a^3+3b^3+c^3)^2} \le 5$$$$\frac{(a^2+2b^2+c^2)^3}{(a^3+3b^3+c^3)^2} \le\frac{29}{9}$$$$\frac{(a^2+3b^2+c^2)^3}{(a^3+2b^3+c^3)^2} \le\frac{35}{4}$$$$\frac{(a^2+3b^2+3c^2)^3}{(a^3+b^3+c^3)^2} \le 55$$

sqing wrote: Let $ a, b, c$ be positive real numbers. Prove that $$\frac{(a^2+2b^2+c^2)^3}{(a^3+b^3+c^3)^2} \le 10$$$$\frac{(a^2+b^2+c^2)^3}{(a^3+2b^3+c^3)^2} \le\frac{9}{4}$$$$\frac{(a^2+2b^2+c^2)^3}{(a^3+2b^3+c^3)^2} \le 4$$

Attachments:

Replace $x_1, x_2, x_3$ with $a, b, c$. Then, we have $$2 \sum_{cyc} \left( a^6 \right) + 6 \sum_{cyc} \left( a^3b^3 \right) = \sum_{cyc} \left( a^6 + 2a^3b^3 \right) + \sum_{cyc} \left( a^6 + 2a^3c^3 \right) + 2 \sum_{cyc} a^3b^3$$$$\ge \sum_{cyc} \left( 3 \sqrt[3]{a^6 \cdot (a^3b^3)^2} \right) + \sum_{cyc} \left(3 \sqrt[3]{a^6 \cdot (a^3c^3)^2} \right) + 2\left(3 \sqrt[3]{a^3b^3 \cdot b^3c^3 \cdot c^3a^3} \right)$$$$= \sum_{cyc} 3a^4b^2 + \sum_{cyc} 3a^4c^2 + 6a^2b^2c^2.$$Now, it's easy to see $$3(a^3 +b^3 + c^3)^2 = 3(a^6+b^6+c^6) + 3 \left(2 \sum_{cyc} a^3b^3 \right) = \sum_{cyc} a^6 + 2 \sum_{cyc} a^6 + 6 \sum_{cyc} a^3b^3$$$$\ge \sum_{cyc} a^6 + \sum_{cyc} 3a^4b^2 + \sum_{cyc} 3a^4c^2 + 6a^2b^2c^2 = (a^2 + b^2 + c^2)^3$$which clearly finishes. $\blacksquare$