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