Question-105380

Question Number 105380 by Skabetix last updated on 28/Jul/20

Commented by Skabetix last updated on 28/Jul/20

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Answered by 1549442205PVT last updated on 28/Jul/20

Apply {Cauchy}^{'} s inequality for three

positive numners we get :

a^{2} + ab + b^{2} ⩾ 3^{3} \sqrt{a^{2} . ab . b^{2}} = 3 ab \Rightarrow \log_{c} (a^{2} + ab + b^{2}) ⩾ \log_{c} (3 ab) = \log_{c} 3 + \log_{c} a + \log_{c} b

Similarly, \log_{a} (b^{2} + bc + c^{2}) ⩾ \log_{a} 3 + \log_{a} b + \log_{a} c

\log_{b} (c^{2} + ca + a^{2}) ⩾ \log_{b} 3 + \log_{b} c + \log_{b} a

Adding up three above inequalities

we get :

LHS ⩾ \log_{a} 3 + \log_{b} 3 + \log_{c} 3 + (\log_{a} b + \log_{b} a) + (\log_{b} c + \log_{c} b) + (\log_{c} a + \log_{a} c)

⩾ \log_{a} 3 + \log_{b} 3 + \log_{c} 3 + 6 (1) (since \log_{x} y + \log_{y} x ⩾ 2 \sqrt{\log_{x} y . \log_{y} x} = 2 due to \log_{x} y . \log_{y} x = 1)

From the hypothesis a . b . c = 3 we get

1 = \log_{3} 3 = \log_{3} (a . b . c) = \log_{3} a + \log_{3} b + \log_{3} c

Therefeore, apply the inequality (x + y + z) (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) ⩾ 9 (x, y, z > 0) we get :

\log_{a} 3 + \log_{b} 3 + \log_{c} 3 ⩾ \frac{9}{\log_{3} a + \log_{3} b + \log_{3} c} = 9 (2)

From the inequalities (1) and (2) we obtain :

LHS ⩾ 9 + 6 = 15 (q . e . d)

The equality ocurrs if and only if

a = b = c =^{3} \sqrt{3}