Let a_1, a_2, a_3 be any positive real number | vedclass

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(D) We use the $AM \ge GM$ inequality for positive real numbers.For option $A$: By $AM \ge GM$,$\frac{a_1^3 + a_2^3 + a_3^3}{3} \ge \sqrt[3]{a_1^3 a_2

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$(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^3 \le 27$

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(D) We use the $AM \ge GM$ inequality for positive real numbers.For option $A$: By $AM \ge GM$,$\frac{a_1^3 + a_2^3 + a_3^3}{3} \ge \sqrt[3]{a_1^3 a_2^3 a_3^3} = a_1 a_2 a_3$,so $a_1^3 + a_2^3 + a_3^3 \ge 3a_1 a_2 a_3$. This is true.For option $B$: By $AM \ge GM$,$\frac{a_1/a_2 + a_2/a_3 + a_3/a_1}{3} \ge \sqrt[3]{\frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdot \frac{a_3}{a_1}} = 1$,so $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} \ge 3$. This is true.For option $C$: By $AM \ge HM$,$(a_1 + a_2 + a_3) \ge \frac{9}{1/a_1 + 1/a_2 + 1/a_3}$,which implies $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} ight) \ge 9$. This is true.For option $D$: The inequality $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} ight)^3 \le 27$ is generally false. For example,if $a_1=a_2=a_3=1$,then $(3)(3^3) = 81 
ot\le 27$. Thus,option $D$ is not true.

Let $a_1, a_2, a_3$ be any positive real numbers,then which of the following statements is not true?

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