Given-a-b-c-R-3-such-that-abc-1-Show-that-a-1-1-b-b-1-1-c-c-1-1-a-1-

Question Number 117122 by Ar Brandon last updated on 09/Oct/20

Given a, b, c \in R^{3} such that abc = 1 . Show that :

(a - 1 + \frac{1}{b}) (b - 1 + \frac{1}{c}) (c - 1 + \frac{1}{a}) ⩽ 1

Commented by MJS_new last updated on 09/Oct/20

I {don}^{'} t think {it}^{'} s true for all (a ∣ b ∣ c) \in R^{3}

Commented by MJS_new last updated on 09/Oct/20

let b = - a \land c = \frac{1}{a b} = - \frac{1}{a^{2}}

\Rightarrow

\frac{(a^{2} + a + 1) (a^{2} - a + 1) (a^{2} - a + 1)}{a^{3}} ⩽ 1 (1)

\frac{a^{6} - a^{5} - 2 a^{3} - a - 1}{a^{3}} ⩽ 0

but the numerator has got 2 real zeros

a_{1} \approx - . 560769 \land a_{2} = - \frac{1}{a_{1}} \approx 1.78326

\Rightarrow (1) is true for a ⩽ a_{1} \lor 0 < a ⩽ a_{2} and

wrong for a_{1} < a < 0 \lor a > a_{2}

Commented by 1549442205PVT last updated on 10/Oct/20

For a = b = - 3, c = 1 / 9 then

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

Since L . H . S = (- 4 - \frac{1}{3}) (- 4 + 9) (\frac{1}{9} - 1 - \frac{1}{3})

= \frac{- 13}{3} \times 5 \times (\frac{- 11}{9}) = \frac{715}{27} > 1

Commented by Ar Brandon last updated on 13/Oct/20

OK Sir, thanks for your suggestions