An Easy Way And A Hard Way II

Let the expression $=k$ . We can rewrite $k$ as:

$\frac{9a(a+b)}{b(a+c)}+\frac{6b(b+c)}{c(a+b)}+\frac{4c(a+c)}{a(b+c)}=k$

Since, $a,b,c>0$ , any combinations of sums, product, and quotients of $a,b,c$ will be positive, so we can use the AM-GM inequality. This gets us:

$\frac{k}{3} \geq \sqrt[3]{\frac{9(a)(a+b)(6)(b)(b+c)(4)(c)(a+c)}{b(a+c)(c)(a+b)(a)(b+c)}}$ .

Everything under the cube root cancels except for $9 \times 6 \times 4$ , so we end up with:

$\frac{k}{3} \geq \sqrt[\large 3]{216}$

Solving for $k$ gives:

$\large k \geq 18$

so our answer is

$\large \boxed{18}$ .

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I deleted the extension of the solution in order to prevent people from instantly getting the answer to my new problem, A Hard Way And A Hard Way