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Question Number 201477 by York12 last updated on 07/Dec/23
how to prove that  (3d_3 +4d_2 +3d_1 )^2 ≤5(d_1 ^2 +d_2 ^2 +d_3 ^2 +(d_2 +d_1 )^2 +(d_3 +d_2 )^2 +(d_1 +d_2 +d_3 )^2 )
$$\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\left(\mathrm{3d}_{\mathrm{3}} +\mathrm{4d}_{\mathrm{2}} +\mathrm{3d}_{\mathrm{1}} \right)^{\mathrm{2}} \leqslant\mathrm{5}\left(\mathrm{d}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{d}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{d}_{\mathrm{3}} ^{\mathrm{2}} +\left(\mathrm{d}_{\mathrm{2}} +\mathrm{d}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\mathrm{d}_{\mathrm{3}} +\mathrm{d}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{d}_{\mathrm{1}} +\mathrm{d}_{\mathrm{2}} +\mathrm{d}_{\mathrm{3}} \right)^{\mathrm{2}} \right) \\ $$
Answered by AST last updated on 07/Dec/23
Let d_1 =a;d_2 =b;d_3 =c  5(3a^2 +4b^2 +3c^2 +4ab+4bc+2ca)≥^?   9a^2 +16b^2 +9c^2 +24bc+24ab+18ac  ⇔3a^2 +2b^2 +3c^2 ≥2ab+2bc+4ac  a^2 +b^2 ≥2ab;2a^2 +2c^2 ≥4ac;b^2 +c^2 ≥2bc  ⇒3a^2 +2b^2 +c^2 ≥2ab+2bc+4ac⇒original inequality  is true.
$${Let}\:{d}_{\mathrm{1}} ={a};{d}_{\mathrm{2}} ={b};{d}_{\mathrm{3}} ={c} \\ $$$$\mathrm{5}\left(\mathrm{3}{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} +\mathrm{4}{ab}+\mathrm{4}{bc}+\mathrm{2}{ca}\right)\overset{?} {\geqslant} \\ $$$$\mathrm{9}{a}^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} +\mathrm{9}{c}^{\mathrm{2}} +\mathrm{24}{bc}+\mathrm{24}{ab}+\mathrm{18}{ac} \\ $$$$\Leftrightarrow\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} \geqslant\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{4}{ac} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{2}{ab};\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} \geqslant\mathrm{4}{ac};{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{2}{bc} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{4}{ac}\Rightarrow{original}\:{inequality} \\ $$$${is}\:{true}. \\ $$
Commented by York12 last updated on 07/Dec/23
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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