Question Number 59131 by naka3546 last updated on 05/May/19
$${a},\:{b},\:{c}\:\:\in\:\:\mathbb{R} \\ $$$${a}\:+\:{b}\:+\:{c}\:\:=\:\:\mathrm{5} \\ $$$${Prove}\:\:{that} \\ $$$$\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:−\:\mathrm{2}{b}\:+\:\mathrm{1}}\:\:+\:\:\sqrt{{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:\mathrm{2}{c}\:+\:\mathrm{1}}\:\:+\:\:\sqrt{{c}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} \:−\:\mathrm{2}{a}\:+\:\mathrm{1}}\:\:\:\geqslant\:\:\sqrt{\mathrm{29}} \\ $$
Answered by Senior Sun last updated on 06/May/19
$${By}\:{Minkoski} \\ $$$$\underset{{cyc}} {\sum}\sqrt{{a}^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} }\geqslant\sqrt{\left(\underset{{cyc}} {\sum}{a}\right)^{\mathrm{2}} +\left(\underset{{cyc}} {\sum}{b}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\sqrt{\mathrm{29}} \\ $$$${there}\:{is}\:{no}\:{equality},\:{so}\:{the}\:{givten}\:{inequality} \\ $$$${is}\:>\sqrt{\mathrm{29}} \\ $$
Answered by tanmay last updated on 05/May/19
![(√(a^2 +(b−1)^2 )) +(√(b^2 +(c−1)^2 )) +(√(c^2 +(a−1)^2 )) ((a^2 +(b−1)^2 )/2)≥[a^2 ×(b−1)^2 ]^(1/2) (√(a^2 +(b−1)^2 )) ≥(√(2(a)(b−1))) given exptession≥(√2) [(√(a(b−1))) +(√(b(c−1))) +(√(c(a−1))) ] (√2)×((a+b−1)/2)≥[a(b−1)]^(1/2) ×(√2) (√2)×((b+c−1)/2)≥[b(c−1)]^(1/2) ×(√2) (√2)×((c+a−1)/2)≥[c(a−1)]^(1/2) ×(√2) (√2)×[((2(a+b+c)−3)/2)]≥(√2) [(√(a(b−1)+b(c−1)+c(a−1))) (√2) ×(7/2)≥D.E [D.E=derived expression]](https://www.tinkutara.com/question/Q59137.png)
$$\sqrt{{a}^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} }\:+\sqrt{{b}^{\mathrm{2}} +\left({c}−\mathrm{1}\right)^{\mathrm{2}} \:}\:+\sqrt{{c}^{\mathrm{2}} +\left({a}−\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\frac{{a}^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}\geqslant\left[{a}^{\mathrm{2}} ×\left({b}−\mathrm{1}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\left({b}−\mathrm{1}\right)^{\mathrm{2}} }\:\geqslant\sqrt{\mathrm{2}\left({a}\right)\left({b}−\mathrm{1}\right)}\: \\ $$$${given}\:{exptession}\geqslant\sqrt{\mathrm{2}}\:\left[\sqrt{{a}\left({b}−\mathrm{1}\right)}\:+\sqrt{{b}\left({c}−\mathrm{1}\right)}\:+\sqrt{{c}\left({a}−\mathrm{1}\right)}\:\right] \\ $$$$\sqrt{\mathrm{2}}×\frac{{a}+{b}−\mathrm{1}}{\mathrm{2}}\geqslant\left[{a}\left({b}−\mathrm{1}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} ×\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}}×\frac{{b}+{c}−\mathrm{1}}{\mathrm{2}}\geqslant\left[{b}\left({c}−\mathrm{1}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} ×\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}}×\frac{{c}+{a}−\mathrm{1}}{\mathrm{2}}\geqslant\left[{c}\left({a}−\mathrm{1}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} ×\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}}×\left[\frac{\mathrm{2}\left({a}+{b}+{c}\right)−\mathrm{3}}{\mathrm{2}}\right]\geqslant\sqrt{\mathrm{2}}\:\left[\sqrt{{a}\left({b}−\mathrm{1}\right)+{b}\left({c}−\mathrm{1}\right)+{c}\left({a}−\mathrm{1}\right)}\:\right. \\ $$$$\sqrt{\mathrm{2}}\:×\frac{\mathrm{7}}{\mathrm{2}}\geqslant{D}.{E}\:\:\left[{D}.{E}={derived}\:{expression}\right] \\ $$$$ \\ $$
Commented by tanmay last updated on 05/May/19
$${to}\:{be}\:{continued}… \\ $$