If a, b, and c are positive real numbers,then | vedclass

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(A) Using the $A.M. \ge G.M.$ inequality for positive real numbers $a, b, c$:$\frac{a + b + c}{3} \ge (abc)^{1/3} \implies a + b + c \ge 3(abc)^{1/3}$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Using the $A.M. \ge G.M.$ inequality for positive real numbers $a, b, c$:$\frac{a + b + c}{3} \ge (abc)^{1/3} \implies a + b + c \ge 3(abc)^{1/3}$Similarly,for $1/a, 1/b, 1/c$:$\frac{1/a + 1/b + 1/c}{3} \ge \left(\frac{1}{abc}ight)^{1/3} \implies \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 3\left(\frac{1}{abc}ight)^{1/3}$Multiplying these two inequalities:$(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}ight) \ge 3(abc)^{1/3} 	imes 3\left(\frac{1}{abc}ight)^{1/3}$$(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}ight) \ge 9$The minimum value is $9$,which occurs when $a = b = c$.

If $a, b,$ and $c$ are positive real numbers,then the minimum value of $(a + b + c)(1/a + 1/b + 1/c)$ is .......

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