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Question Number 170050 by DAVONG last updated on 14/May/22
Show that,∀a,b,c∈R :a^2 +b^2 +c^2 +12≥4(a+b+c)
$$\mathrm{Show}\:\mathrm{that},\forall\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}\::\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{12}\geqslant\mathrm{4}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$
Answered by mindispower last updated on 15/May/22
12=4+4+4  4+x^2 ≥4x⇒  a^2 +4+b^2 +4+c^2 +4≥4a+4b+4c=4(a+b+c)
$$\mathrm{12}=\mathrm{4}+\mathrm{4}+\mathrm{4} \\ $$$$\mathrm{4}+{x}^{\mathrm{2}} \geqslant\mathrm{4}{x}\Rightarrow \\ $$$${a}^{\mathrm{2}} +\mathrm{4}+{b}^{\mathrm{2}} +\mathrm{4}+{c}^{\mathrm{2}} +\mathrm{4}\geqslant\mathrm{4}{a}+\mathrm{4}{b}+\mathrm{4}{c}=\mathrm{4}\left({a}+{b}+{c}\right) \\ $$

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