Question Number 201615 by hardmath last updated on 09/Dec/23
$$\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}}\:+\:\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{c}}\:\geqslant\:\frac{\left(\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\right)^{\mathrm{2}} }{\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}} \\ $$
Answered by AST last updated on 09/Dec/23
$$\sqrt{\left(\frac{{x}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\:\sqrt{{b}}}\right)^{\mathrm{2}} +\left(\frac{{z}}{\:\sqrt{{c}}}\right)^{\mathrm{2}} }\sqrt{\left(\sqrt{{a}}\right)^{\mathrm{2}} +\left(\sqrt{{b}}\right)^{\mathrm{2}} +\left(\sqrt{{c}}\right)^{\mathrm{2}} } \\ $$$$\geqslant\left({x}+{y}+{z}\right) \\ $$$$\mid{u}\mid\mid{v}\mid\geqslant{u}\centerdot{v}\:{where}\:{u}=\langle\frac{{x}}{\:\sqrt{{a}}},\frac{{y}}{\:\sqrt{{b}}},\frac{{z}}{\:\sqrt{{c}}}\rangle;{v}=\langle\sqrt{{a}},\sqrt{{b}},\sqrt{{c}}\rangle \\ $$