If a - b - c -1- where a - b - c are positive real numbers- then the minimum value of 1- ab -1- bc -1- ca isA- 13B- 24C -27D- 32

The correct option is C 271ab+1bc+1ac =a+b+cabc=1abc We know that, A.M. ≥ G.M ⇒a+b+c3≥ (abc)^1/3 ⇒( 1 3)^3 ≥ abc ⇒1abc≥ 27 Therefore, 1ab+1bc+1ac≥ 27 Hence, ...

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Standard XII

Mathematics

Arithmetic Mean

If a + b + c ...

Question

If $a + b + c = 1$ , where $a, b, c$ are positive real numbers, then the minimum value of $\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}$ is

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27

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Solution

The correct option is C $27$
$\frac{1}{a b} + \frac{1}{b c} + \frac{1}{a c} = \frac{a + b + c}{a b c} = \frac{1}{a b c}$
We know that,
$A . M . \geq G . M \Rightarrow \frac{a + b + c}{3} \geq (a b c)^{1 / 3} \Rightarrow {(\frac{1}{3})}^{3} \geq a b c \Rightarrow \frac{1}{a b c} \geq 27$

Therefore,
$\frac{1}{a b} + \frac{1}{b c} + \frac{1}{a c} \geq 27$

Hence, the minimum value of the given expression is $27$ .

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If a+b+c=1, where a,b,c are positive real numbers, then the minimum value of 1ab+1bc+1ca is

The correct option is C 271ab+1bc+1ac=a+b+cabc=1abc We know that, A.M.≥G.M⇒a+b+c3≥(abc)1/3⇒(13)3≥abc⇒1abc≥27 Therefore, 1ab+1bc+1ac≥27 Hence, the minimum value of the given expression is 27.

If $a + b + c = 1$ , where $a, b, c$ are positive real numbers, then the minimum value of $\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}$ is