Question Number 83965 by Rio Michael last updated on 08/Mar/20
$$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove}\left(\mathrm{with}\:\mathrm{counter}−\mathrm{example}\right)\:\mathrm{that} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{For}\:\mathrm{all}\:\mathrm{two}\:\mathrm{dimensional}\:\mathrm{vectors}\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}, \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{b}}\:=\:\boldsymbol{\mathrm{a}}.\:\boldsymbol{\mathrm{c}}\:\Rightarrow\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{For}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{a}\:+{b}}{\mathrm{2}}\:\geqslant\:\sqrt{{ab}}\: \\ $$
Commented by mr W last updated on 08/Mar/20
$$\left({a}\right) \\ $$$${statement}\:{is}\:{wrong}! \\ $$$$\boldsymbol{{a}}\centerdot\boldsymbol{{b}}=\mid{a}\mid\mid{b}\mid\mathrm{cos}\:\beta \\ $$$$\boldsymbol{{a}}\centerdot\boldsymbol{{c}}=\mid{a}\mid\mid{c}\mid\mathrm{cos}\:\gamma \\ $$$$\boldsymbol{{a}}\centerdot\boldsymbol{{b}}=\boldsymbol{{a}}\centerdot\boldsymbol{{c}}\:\Rightarrow\mid{b}\mid\:\mathrm{cos}\:\beta=\mid{c}\mid\:\mathrm{cos}\:\gamma\:\nRightarrow\:\boldsymbol{{b}}=\boldsymbol{{c}} \\ $$$${example} \\ $$$${a}=\left(\mathrm{1},\mathrm{0}\right) \\ $$$${b}=\left(\mathrm{1},\mathrm{1}\right) \\ $$$${c}=\left(\mathrm{1},\mathrm{2}\right) \\ $$$${a}\centerdot{b}={a}\centerdot{c}=\mathrm{1} \\ $$$${but}\:{b}\neq{c} \\ $$
Commented by mr W last updated on 08/Mar/20
$$\left({b}\right) \\ $$$${for}\:{a},{b}>\mathrm{0}: \\ $$$$\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow{a}−\mathrm{2}\sqrt{{ab}}+{b}\geqslant\mathrm{0} \\ $$$$\Rightarrow\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$
Commented by Rio Michael last updated on 08/Mar/20
$${perfect}\:{sir},\:{and}\:{thanks}\:{for}\:{the}\:{correction} \\ $$