Question

If a, b, c are real and positive then what is the minimum value of

$\frac{( {a}^{2} + a + 1)( {b}^{2} + b + 1)( {c}^{2} +c +1) }{abc}$

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Answer 1

$\green{\large\underline{\sf{Solution-}}}$

Given that

$\rm :\longmapsto\:a > 0, \: b > 0, \: c > 0$

and

The expression is

$\rm :\longmapsto\:\dfrac{( {a}^{2} + a + 1)( {b}^{2} + b + 1)( {c}^{2} +c +1) }{abc}$

can be rewritten as

$\rm \:  =  \: \bigg[\dfrac{ {a}^{2} + a + 1}{a} \bigg]\bigg[\dfrac{ {b}^{2} + b + 1}{b} \bigg]\bigg[\dfrac{ {c}^{2} + c + 1}{c} \bigg]$

$\rm \:  =  \: \bigg[a + 1 + \dfrac{1}{a} \bigg]\bigg[b + 1 + \dfrac{1}{b} \bigg]\bigg[c + 1 + \dfrac{1}{c} \bigg]$

can be further rewritten as

$\rm \:  =  \: \bigg[a + \dfrac{1}{a} + 1 \bigg]\bigg[b + \dfrac{1}{b} + 1 \bigg]\bigg[c + \dfrac{1}{c} + 1\bigg]$

Now, we know that

$\boxed{\tt{ \: \: \: x \: \: + \: \: \frac{1}{x} \: \: \geqslant \: \: 2 \: \: \: \: \: \: if \: \: \: x > 0 \: \: \: }}$

So, it means,

$\red{\rm :\longmapsto\:a + \dfrac{1}{a} \geqslant 2}$

$\red{\rm :\longmapsto\:b + \dfrac{1}{b} \geqslant 2}$

$\red{\rm :\longmapsto\:c + \dfrac{1}{c} \geqslant 2}$

So, on substituting these values in above expression,

$\rm \:  \geqslant   \: (2 + 1)(2 + 1)(2 + 1)$

$\rm \:  \geqslant   \: 3 \times 3 \times 3$

$\rm \:  \geqslant   \: 27$

Hence,

$\green{\rm \implies\:\boxed{\sf{ \dfrac{( {a}^{2} + a + 1)( {b}^{2} + b + 1)( {c}^{2} +c +1) }{abc} \geqslant 27 \: }}}$

So,

Minimum value

$\red{\rm \implies\:\boxed{\sf{ \dfrac{( {a}^{2} + a + 1)( {b}^{2} + b + 1)( {c}^{2} +c +1) }{abc} = 27 \: }}}$

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To prove that

$\boxed{\tt{ \: \: \: x \: \: + \: \: \frac{1}{x} \: \: \geqslant \: \: 2 \: \: \: \: \: \: if \: \: \: x > 0 \: \: \: }}$

PROOF :-

Let us consider two positive real numbers x and 1/x.

We know, Arithmetic mean AM and Geometric mean GM are connected by the relationship

$\red{\rm :\longmapsto\:AM \geqslant GM}$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg[x + \dfrac{1}{x} \bigg] \geqslant \sqrt{x \times \dfrac{1}{x} }$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg[x + \dfrac{1}{x} \bigg] \geqslant \sqrt{1 }$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg[x + \dfrac{1}{x} \bigg] \geqslant 1$

$\bf\implies \:x + \dfrac{1}{x} \geqslant 2$

HENCE PROVED