a, b, c ∈ R a + b + c = 5 Prove that (√(a^2 + b^2 − 2b + 1)) + (√(b^2 + c^2 − 2c + 1)) + (√(c^2 + a^2 − 2a + 1)) ≥ (√(29))


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Question Number 59131 by naka3546 last updated on 05/May/19

$a, b, c \in R$ $a + b + c = 5$ $P r o v e t h a t$ $\sqrt{a^{2} + b^{2} - 2 b + 1} + \sqrt{b^{2} + c^{2} - 2 c + 1} + \sqrt{c^{2} + a^{2} - 2 a + 1} ⩾ \sqrt{29}$
	Answered by Senior Sun last updated on 06/May/19

	$B y M i n k o s k i$ $\sum_{c y c} \sqrt{a^{2} + {(b - 1)}^{2}} ⩾ \sqrt{{(\sum_{c y c} a)}^{2} + {(\sum_{c y c} b - 3)}^{2}}$ $⩾ \sqrt{5^{2} + {(5 - 3)}^{2}}$ $⩾ \sqrt{29}$ $t h e r e i s n o e q u a l i t y, s o t h e g i v t e n i n e q u a l i t y$ $i s > \sqrt{29}$
	Answered by tanmay last updated on 05/May/19

	$\sqrt{a^{2} + {(b - 1)}^{2}} + \sqrt{b^{2} + {(c - 1)}^{2}} + \sqrt{c^{2} + {(a - 1)}^{2}}$ $\frac{a^{2} + {(b - 1)}^{2}}{2} ⩾ {[a^{2} \times {(b - 1)}^{2}]}^{\frac{1}{2}}$ $\sqrt{a^{2} + {(b - 1)}^{2}} ⩾ \sqrt{2 (a) (b - 1)}$ $g i v e n e x p t e s s i o n ⩾ \sqrt{2} [\sqrt{a (b - 1)} + \sqrt{b (c - 1)} + \sqrt{c (a - 1)}]$ $\sqrt{2} \times \frac{a + b - 1}{2} ⩾ {[a (b - 1)]}^{\frac{1}{2}} \times \sqrt{2}$ $\sqrt{2} \times \frac{b + c - 1}{2} ⩾ {[b (c - 1)]}^{\frac{1}{2}} \times \sqrt{2}$ $\sqrt{2} \times \frac{c + a - 1}{2} ⩾ {[c (a - 1)]}^{\frac{1}{2}} \times \sqrt{2}$ $\sqrt{2} \times [\frac{2 (a + b + c) - 3}{2}] ⩾ \sqrt{2} [\sqrt{a (b - 1) + b (c - 1) + c (a - 1)}$ $\sqrt{2} \times \frac{7}{2} ⩾ D . E [D . E = d e r i v e d e x p r e s s i o n]$
	Commented by tanmay last updated on 05/May/19

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