Question

If a, b, c are real distinct positive numbers such that a + b + c = 1, prove that

$\frac{(1 + a)(1 + b)(1 + c)}{(1 - a)(1 - b)(1 - c)} > 8$
​

Answer 1

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

Given that

$\purple{\rm :\longmapsto\:a \: \ne \: b \: \ne \: c \: > \: 0}$

And

$\rm :\longmapsto\:a + b + c = 1$

can be rewritten as

$\rm :\longmapsto\:a = 1 - b - c$

On adding 1 on both sides, we get

$\rm :\longmapsto\:1 + a = 1 + 1 - b - c$

can be re-arranged as

$\bf :\longmapsto\:1 + a = 1 - b + 1 - c - - - (1)$

Now, we know relationship between Arithmetic mean (AM) and Geometric mean (GM)

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: AM \: \: \geqslant \: \: GM \: \: }}}$

Now, (1 - b) and (1 - c) are two distinct real positive numbers so,

$\rm :\longmapsto\:\dfrac{1 - b + 1 - c}{2} > \sqrt{(1 - b)(1 - c)}$

$\bf\implies \:(1 - b) + (1 - c) > 2 \sqrt{(1 - b)(1 - c)}$

On substituting this value in equation (1), we get

$\bf\implies \:1 + a > 2 \sqrt{(1 - b)(1 - c)} - - (2)$

Similarly,

$\bf\implies \:1 + b > 2 \sqrt{(1 - a)(1 - c)} - - (3)$

and

$\bf\implies \:1 + c > 2 \sqrt{(1 - a)(1 - b)} - - (4)$

On multiply equation (2), (3) and (4), we get

$\rm :\longmapsto\:(1 + a)(1 + b)(1 + c) > 8(1 - a)(1 - b)(1 - c)$

$\purple{\rm\implies \:\boxed{\tt{ \: \: \frac{(1 + a)(1 + b)(1 + c)}{(1 - a)(1 - b)(1 - c)} > 8 \: \: }}}$

Hence, Proved

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EXPLORE MORE

If and b are positive real numbers, then

1. Arithmetic mean (AM) between a and b is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ AM = \frac{a + b}{2}}}}$

2. Geometric mean (GM) between a and b is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: GM \: = \: \sqrt{ab} \: \: }}}$

3. Harmonic mean (HM) between a and b is given by

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: \: HM = \frac{2ab}{a + b} \: \: \: }}}$

4. Relationship between Arithmetic mean, Geometric mean and Harmonic mean

$\boxed{\tt{ AM \geqslant GM \geqslant HM}} \\ \\ \boxed{\tt{ {GM}^{2} = AM \times HM}}$