Question 711123
First you multiply and simplify.
I would also rearrange as shown below.
{{{(a + b + c)(1/a + 1/b + 1/c)=a(1/a)+a/b+a/c+b/a+b(1/b)+b/c+c/a+c/b+c(1/c)=1+a/b+a/c+b/a+1+b/c+c/a+c/b+1=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)}}}
Next we have to show that each expression in brackets is equal or greater than 2.
All of them are of the form
{{{x/y+y/x}}}
We could probe it with {{{x/y+y/x}}} or with {{{a/b+b/a}}}.
{{{x/y+y/x=(x^2+y^2)/xy}}}
We want to prove that {{{(x^2+y^2)/xy>=2}}}
{{{(x^2+y^2)-2xy=(x-y)^2}}}
so {{{(x^2+y^2)-2xy>=0}}} --> {{{(x^2+y^2)>=2xy}}}
And since x and y are positive xy is a positive number we can use to divide both sides of the inequality to find that
{{{(x^2+y^2)/xy>=2xy/xy}}} --> {{{(x^2+y^2)/xy>=2}}} --> {{{x/y+y/x>=2}}}
Then
{{{a/b+b/a>=2}}} ,
{{{a/c+c/a>=2}}} and
{{{b/c+c/b>=2}}}
So
{{{(a + b + c)(1/a + 1/b + 1/c)=3+(a/b+b/a)+(a/c+c/a)+(b/c+c/b)>=3+2+2+2}}}
{{{(a + b + c)(1/a + 1/b + 1/c)>=9}}}