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AM-GM inequality question

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Let $$a,b,c,d$$   be positive real numbers. Find the minimum value of $$\frac{(a + b)(a + c)(b + c)}{abc}.$$

I think you're supposed to use AM-GM inequalities, but I have no idea how. Help!

Apr 20, 2020

#1
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The minimum value is 4.

Apr 21, 2020
#2
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Let $$a,b,c,d$$ be positive real numbers.

Find the minimum value of $$\dfrac{(a + b)(a + c)(b + c)}{abc}$$

$$\mathbf{\huge{AM \geq GM }}$$

$$\begin{array}{|rcll|} \hline \dfrac{a+b}{2} & \ge & \sqrt{ab} \\\\ \dfrac{a+c}{2} & \ge & \sqrt{ac} \\\\ \dfrac{b+c}{2} & \ge & \sqrt{bc} \\\\ \hline \left( \dfrac{a+b}{2} \right) \left( \dfrac{a+c}{2} \right) \left( \dfrac{b+c}{2} \right) & \ge & \sqrt{ab}\sqrt{ac} \sqrt{bc} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & \sqrt{abacbc} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & \sqrt{a^2b^2c^2} \\\\ \dfrac{(a + b)(a + c)(b + c)}{8} & \ge & abc \\\\ \dfrac{(a + b)(a + c)(b + c)}{abc} & \ge & 8 \\\\ \mathbf{ 8 } & \le & \mathbf{\dfrac{(a + b)(a + c)(b + c)}{abc}} \\ \hline \end{array}$$

The minimum value is 8

Apr 21, 2020
#3
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thanks so much, heureka!!!! it was correct!

Guest Apr 23, 2020