Use the A.M.-G.M Inequality to prove the following statement

[shivers20]

I am confused how they used the A.M. G.M. inequaltiy to solve this proof. Would anyone care to explain.

Let a,b,c be Real numbers. If a, b, c >= 0, then (a + b)(a + c)(b + c) >= 8abc

Answer:
The A.M. - G.M. inequality states that for a set of non-negative real
numbers, the arithmetic mean is greater than or equal to the the geometric
mean with the equality holding only if all numbers in the set are equal.
For real numbers a, b, c >= 0 we can use the A.M. - G.M. inequality to write
three equations:

a+b/2 >= sqrt[ab]

a+c/2 >= sqrt[ac]

b+c/2 >= sqrt[bc]


Multiply the left-hand side of the equation by the left-hand side of equation
and the right-hand side of the equation by the right-hand side of
equation to get:
((a + b)/2)) x ((a+c)/2)) >=sqrt [ab] x sqrt [ac]

(a+b)(a+c)(b+c)/8 >= sqrt[(ab)(ac)(bc)

Finally, simplify sqrt[a^2 b^2 c^2] = abc and multiply both sides sides by 8.

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
\left( {\frac{{a + b}}{2}} \right)\left( {\frac{{a + c}}{2}} \right)\left( {\frac{{b + c}}{2}} \right) & \ge & \left( {\sqrt {ab} } \right)\left( {\sqrt {ac} } \right)\left( {\sqrt {bc} } \right) \\
& \ge & 8\sqrt {a^2 b^2 c^2 } \\
& \ge & 8abc \\
\end{array}\)