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

Answered by gunawan last updated on 20/Jun/18

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

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

$$excellent...$$

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