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

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(D) We use the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,which states that for positive real numbers,$AM \ge GM \ge HM$.$1$. For option $(A)

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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 Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,which states that for positive real numbers,$AM \ge GM \ge HM$.$1$. For option $(A)$: By $AM \ge GM$,we have $\frac{a_1^3 + a_2^3 + a_3^3}{3} \ge \sqrt[3]{a_1^3 a_2^3 a_3^3} = a_1 a_2 a_3$,which implies $a_1^3 + a_2^3 + a_3^3 \ge 3a_1 a_2 a_3$. This is true.$2$. For option $(B)$: By $AM \ge GM$,we have $\frac{\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1}}{3} \ge \sqrt[3]{\frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdot \frac{a_3}{a_1}} = \sqrt[3]{1} = 1$,which implies $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} \ge 3$. This is true.$3$. For option $(C)$: By $AM \ge HM$,we have $\frac{a_1 + a_2 + a_3}{3} \ge \frac{3}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}}$,which implies $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) \ge 9$. This is true.$4$. For option $(D)$: Since $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) \ge 9$,the expression $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^3$ will generally be $\ge 9 \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^2$,which is not necessarily $\le 27$. Thus,statement $(D)$ 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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