how to prove?

Question

If abc(a+b+c)=3, prove that (a+b)(b+c)(c+a)≥8

Answer 1

We apply AM$\geq$GM

$\frac{a+b}{2}\geq \sqrt{ab}$      ...1

$\frac{b+c}{2}\geq \sqrt{bc}$      ...2

$\frac{a+c}{2}\geq \sqrt{ac}$      ...3

Multiply 1,2,3

$\frac{(a+b)(b+c)(c+a)}{8}\geq abc$

$(a+b)(b+c)(c+a)\geq 8abc$

Now we have to prove that abc$\leq 1$ for equation to hold

again we apply AM$\geq$GM

$\frac{a+b+c}{3}\geq abc^{\frac{1}{3}}$   ...4

we have abc(a+b+c)=3

$(a+b+c)=\frac{3}{abc}$

substituting above result in equation 4 we get

$\frac{3}{\frac{abc}{3}}\geq abc^{\frac{1}{3}}$

$1\geq abc^{\frac{4}{3}}$

hence if $1\geq abc^{\frac{4}{3}}$ then $abc \leq 1$