Question Number 194648 by universe last updated on 12/Jul/23
Commented by Frix last updated on 12/Jul/23
$$\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Commented by universe last updated on 12/Jul/23
$${solution}\:? \\ $$
Commented by Frix last updated on 12/Jul/23
$${a}={b}={c}\wedge{d}={e}={f} \\ $$
Answered by witcher3 last updated on 12/Jul/23
$$\left(\mathrm{ad}+\mathrm{be}+\mathrm{cf}\right)\leqslant\sqrt{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{d}^{\mathrm{2}} +\mathrm{e}^{\mathrm{2}} +\mathrm{f}^{\mathrm{2}} \right)} \\ $$$$\mathrm{cachy}\:\mathrm{shwartz}\: \\ $$$$\mathrm{use}\:\mathrm{when}\:\mathrm{we}\:\mathrm{have}\:\mathrm{equality} \\ $$
Answered by deleteduser1 last updated on 12/Jul/23
$${ad}+{be}+{cf}\leqslant\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\sqrt{{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} }=\mathrm{56} \\ $$$${Equality}\:{holds}\:{when}\:{a}={kd},{b}={ke},{c}={kf} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{d}+{e}+{f}}={k} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{64}={k}^{\mathrm{2}} \left({d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)=\mathrm{49}{k}^{\mathrm{2}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{8}}{\mathrm{7}}=\frac{{a}+{b}+{c}}{{d}+{e}+{f}}\Rightarrow\frac{{d}+{ef}}{{a}+{b}+{c}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Answered by mr W last updated on 14/Jul/23
$$\boldsymbol{{p}}=\left({a},{b},{c}\right) \\ $$$$\boldsymbol{{q}}=\left({d},{e},{f}\right) \\ $$$$\mid\boldsymbol{{p}}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\sqrt{\mathrm{64}}=\mathrm{8} \\ $$$$\mid\boldsymbol{{q}}\mid=\sqrt{{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} }=\sqrt{\mathrm{49}}=\mathrm{7} \\ $$$${ad}+{be}+{cf}=\boldsymbol{{p}}\centerdot\boldsymbol{{q}}=\mid\boldsymbol{{p}}\mid\mid\boldsymbol{{q}}\mid\mathrm{cos}\:\theta \\ $$$$\mathrm{56}=\mathrm{8}×\mathrm{7}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\mathrm{1}\:\Rightarrow\theta=\mathrm{0}°\:\Rightarrow\boldsymbol{{p}}//\boldsymbol{{q}} \\ $$$$\Rightarrow\frac{{d}}{{a}}=\frac{{e}}{{b}}=\frac{{f}}{{c}}={k}=\pm\frac{\mid\boldsymbol{{q}}\mid}{\mid\boldsymbol{{p}}\mid}=\pm\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow\frac{{d}+{e}+{f}}{{a}+{b}+{c}}={k}=\pm\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Commented by deleteduser1 last updated on 15/Jul/23
$${Observe}\:{that}\:{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} ={k}\left({ad}+{be}+{cf}\right)=\mathrm{56}{k}\geqslant\mathrm{0} \\ $$$$\Rightarrow{k}\:{is}\:{positive} \\ $$
Commented by mr W last updated on 17/Jul/23
$${yes},\:{you}\:{are}\:{right}!\:{thanks}! \\ $$
Commented by kapoorshah last updated on 17/Jul/23
$${nice} \\ $$