Prove that if $a, b, c \geq 0$ then $(a + b)(b + c)( a + c) \geq 8abc$

Question

Prove that if $a, b, c\ge 0$, then $(a + b)(b + c)(a + c)\ge 8abc$.

Answer 1

Note that $$a+b \ge 2 \sqrt{ab}$$ By $\text{AM-GM}$. So we have $$(a+b)(b+c)(c+a) \ge 2\sqrt{ab} \times 2 \sqrt{bc} \times 2 \sqrt{ca}=8abc$$

Answer 2

The ArithmeticMean-GeometricMean (AM-GM) Inequality states that for non-negative real numbers $a$ and $b$ one has:

$$\frac{a+b}{2}\geq \sqrt{a\cdot b}$$

Proof:

$0\leq^{\color{gray}{\star}} (\sqrt{a}-\sqrt{b})^2 = (\sqrt{a})^2-2\sqrt{a}\cdot\sqrt{b}+(\sqrt{b})^2 = a-2\sqrt{ab}+b$

The inequality marked $\star$ is valid since the square of any real number is non-negative (noting that $\sqrt{a}-\sqrt{b}$ is a real number).

Rearranging: $2\sqrt{ab}\leq a+b$

Using this inequality three times (once for $(a+b)$, again for $(b+c)$ and once more for $(c+a)$) the original problem is proven.

Answer 3

Observe that $LHS - RHS = a(b-c)^2 + b(c-a)^2 + c(a-b)^2 \ge 0$.