Application of the arithmetic mean - geometric mean inequality

Use the Arithmetic mean – Geometric Mean inequality to prove that for $a,b,c \in \mathbb{R}_{>0}$ , with $abc =8$ , then

$a+b+c \geq 6 \qquad \text{and} \qquad ab + ac + bc \geq 12.$

Proof. Since $a,b,c$ are positive real numbers from the AM-GM inequality we know,

$\begin{align*} &&G(a,b,c) &\leq M_1 (a,b,c) \\ \implies && (abc)^{1/3} &\leq \left( \frac{a+b+c}{3} \right) \\ \implies && 8 &\leq \frac{(a+b+c)^3}{27} \\ \implies && 27\cdot 8 &\leq (a+b+c)^3 \\ \implies && a+b+c &\geq 6. \end{align*}$

Then, for the second inequality we consider $M_{-1}$ and use the inequality from part (b) of the AM-GM exercise,

$\begin{align*} && M_{-1} (a,b,c) &\leq G(a,b,c) \\ \implies && \left( \frac{a^{-1} + b^{-1} + c^{-1}}{3} \right)^{-1} &\leq (abc)^{1/3} \\ \implies && \left( \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) &\leq 8^{1/3} \\ \implies && \frac{27}{\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^3} &\leq 8 \\ \implies && \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^3 &\geq \frac{27}{8} \\ \implies && \frac{1}{a} + \frac{1}{b} + \frac{1}{c} &\geq \frac{3}{2} \\ \implies && \frac{bc + ac + ab}{abc} &\geq \frac{3}{2} \\ \implies && ab + ac + bc &\geq \frac{3}{2} \cdot 8 = 12. \qquad \blacksquare \end{align*}$

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Application of the arithmetic mean – geometric mean inequality

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