Question Number 185828 by normans last updated on 28/Jan/23
Commented by mr W last updated on 28/Jan/23
$$\left({x}+{y}+{z}\right)_{{min}} =\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\Delta}{\mathrm{2}}} \\ $$$$\Delta={area}\:{of}\:{triangle}\:{ABC} \\ $$$$\mathrm{cos}\:{B}=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:{B}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Delta=\frac{\mathrm{2}×\mathrm{4}×\sqrt{\mathrm{3}}}{\mathrm{2}×\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\left({x}+{y}+{z}\right)_{{min}} =\sqrt{\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}×\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{7}} \\ $$
Commented by mr W last updated on 28/Jan/23
$${see}\:{Q}\mathrm{164174} \\ $$
Commented by normans last updated on 28/Jan/23
$${nice}\:{sir} \\ $$