Can someone tell me what should i do next? Should I use the inequality between Arithmetic and Geometric mean? Problem 1: Let $a,b,c> 0$
$ab+3bc+2ca\leqslant18$
Prove that:
$\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$.
I started on this way:
 $\frac{3}{a} + \frac{2}{b}+ \frac{1}{c}\geqslant 3$
$\frac{3bc+2ac+ab}{abc}\geqslant 3  $
$\frac{ab+3bc+2ac}{abc}\geqslant 3 \times \frac{abc}{3}$
$\frac{ab+3bc+2ca}{3}\geqslant abc$
 A: By the AM-GM inequality, we have $$\left(\frac{ab+3bc+2ac}{3}\right)^3\ge 6{(abc)}^2$$
$$\Rightarrow \frac 1{abc}\ge \sqrt{\frac{3^3.6}{(ab+3bc+2ac)^3}}$$
$$\Rightarrow \frac{ab+3bc+2ac}{abc}\ge \sqrt{\frac {3^3.6}{ab+3bc+2ac}}\ge\sqrt{\frac{3^3.6}{18}}=3$$
A: By Holder:
$$18\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)^2\geq(ab+3bc+2ac)\left(\frac{3}{a}+\frac{2}{b}+\frac{1}{c}\right)\left(\frac{2}{b}+\frac{1}{c}+\frac{3}{a}\right)\geq$$
$$\geq\left(\sqrt[3]{ab\cdot\frac{3}{a}\cdot\frac{2}{b}}+\sqrt[3]{3bc\cdot\frac{2}{b}\cdot\frac{1}{c}}+\sqrt[3]{2ac\cdot\frac{1}{c}\cdot\frac{3}{a}}\right)^3=162$$
and we are done!
The Holder inequality for three sequences is the following.

Let $a_1$, $a_2$,..., $a_n$, $b_1$, $b_2$,..., $b_n$, $c_1$, $c_2$,..., $c_n$,  $\alpha$, $\beta$ and $\gamma$ be positive numbers. Prove that:
  $$(a_1+a_2+...+a_n)^{\alpha}(c_1+c_2+...+c_n)^{\gamma}(b_1+b_2+...+b_n)^{\beta}\geq$$$$\geq\left(\left(a_1^{\alpha}b_1^{\beta}c_1^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+\left(a_2^{\alpha}b_2^{\beta}c_2^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}+...+\left(a_n^{\alpha}b_n^{\beta}c_n^{\gamma}\right)^{\frac{1}{\alpha+\beta+\gamma}}\right)^{\alpha+\beta+\gamma}.$$

It follows from convexity of $f(x)=x^k$, where $k>1$.
In our case $\alpha=\beta=\gamma=1$.
