Question-37964

Question Number 37964 by naka3546 last updated on 20/Jun/18

Answered by gunawan last updated on 20/Jun/18

put (a+b)≥2c (((a+b)(a−b))/c)≥2(a−b) ((a^2 −b^2 )/c)≥2a−2b ..(i) c−b≤0 2a≥b+c then: (c−b)(b+c)≥2a(c−b) ((c^2 −b^2 )/a)≥2c−2b...(ii) a+c≥b (a+c)(a−c)≥b(a−c) ((a^2 −c^2 )/b)≥a−c ...(iii) add (i),(ii), and (iii) so ((a^2 −b^2 )/c)+((c^2 −b^2 )/a)+((a^2 −c^2 )/b)≥(2a+a)+(−2b−2b)+(2c−c) ((a^2 −b^2 )/c)+((c^2 −b^2 )/a)+((a^2 −c^2 )/b)≥3a−4b+c

$$\mathrm{put}\: \\ $$$$\left({a}+{b}\right)\geqslant\mathrm{2}{c} \\ $$$$\frac{\left({a}+{b}\right)\left({a}−{b}\right)}{{c}}\geqslant\mathrm{2}\left({a}−{b}\right) \\ $$$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}}\geqslant\mathrm{2}{a}−\mathrm{2}{b}\:\:\:..\left({i}\right) \\ $$$${c}−{b}\leqslant\mathrm{0} \\ $$$$\mathrm{2}{a}\geqslant{b}+{c}\: \\ $$$$\mathrm{then}: \\ $$$$\left({c}−{b}\right)\left({b}+{c}\right)\geqslant\mathrm{2}{a}\left({c}−{b}\right) \\ $$$$\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}}\geqslant\mathrm{2}{c}−\mathrm{2}{b}…\left({ii}\right) \\ $$$${a}+{c}\geqslant{b} \\ $$$$\left({a}+{c}\right)\left({a}−{c}\right)\geqslant{b}\left({a}−{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{b}}\geqslant{a}−{c}\:\:…\left({iii}\right) \\ $$$${add}\:\left({i}\right),\left({ii}\right),\:\mathrm{and}\:\left({iii}\right)\:\mathrm{so} \\ $$$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}}+\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}}+\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{b}}\geqslant\left(\mathrm{2}{a}+{a}\right)+\left(−\mathrm{2}{b}−\mathrm{2}{b}\right)+\left(\mathrm{2}{c}−{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}}+\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}}+\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{b}}\geqslant\mathrm{3}{a}−\mathrm{4}{b}+{c} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18

$${excellent}… \\ $$

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