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Prove-that-a-2-b-2-c-2-6-3-a-b-c-a-b-c-R-




Question Number 126175 by naka3546 last updated on 17/Dec/20
Prove  that        a^2  + b^2  + c^2  + 6 ≥ 3(a + b + c)  a, b, c ∈ R^+
$${Prove}\:\:{that} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{6}\:\geqslant\:\mathrm{3}\left({a}\:+\:{b}\:+\:{c}\right) \\ $$$${a},\:{b},\:{c}\:\in\:\mathbb{R}^{+} \\ $$
Commented by PRITHWISH SEN 2 last updated on 18/Dec/20
(a −2)^2  ≥−(a−2)  equality holds for a=1  ⇒a^2 +2≥3a  similariy   b^2 +2≥3b   c^2 +2≥3c  adding  a^2 +b^2 +c^2 +6 ≥3(a+b+c)
$$\left(\mathrm{a}\:−\mathrm{2}\right)^{\mathrm{2}} \:\geqslant−\left(\mathrm{a}−\mathrm{2}\right)\:\:\mathrm{equality}\:\mathrm{holds}\:\mathrm{for}\:\mathrm{a}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{2}\geqslant\mathrm{3a} \\ $$$$\mathrm{similariy} \\ $$$$\:\mathrm{b}^{\mathrm{2}} +\mathrm{2}\geqslant\mathrm{3b} \\ $$$$\:\mathrm{c}^{\mathrm{2}} +\mathrm{2}\geqslant\mathrm{3c} \\ $$$$\mathrm{adding} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{6}\:\geqslant\mathrm{3}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$

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