Question Number 107242 by mnjuly1970 last updated on 09/Aug/20
$$\:\:\:\:\:…{question}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:{that}: \\ $$$$\:{if}\:\:{a},{b},{c}\in\mathbb{R}^{+} \:{then}: \\ $$$$\:\:\:\:\:\clubsuit\:\:\:\sqrt{{a}}\:+\sqrt{{b}}+\sqrt{{c}}>\:\sqrt{{a}+{b}+{c}}\:\clubsuit\: \\ $$$$\:\:\:\:\:\:\:….{sincerly}\:{yours}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\mathscr{M}.\mathscr{N}… \\ $$$$\:\: \\ $$
Answered by abdomathmax last updated on 09/Aug/20
$$\mathrm{a}=\:\mathrm{x}^{\mathrm{2}} \:,\mathrm{b}=\mathrm{y}^{\mathrm{2}} \:,\mathrm{c}\:=\mathrm{z}^{\mathrm{2}} \:\:\:\:\left(\mathrm{i}\right)\Rightarrow\mathrm{x}+\mathrm{y}\:+\mathrm{z}>\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} >\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}\:} \:+\mathrm{2}\left(\mathrm{xy}\:+\mathrm{yz}\:+\mathrm{zx}\right)>\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}} \\ $$$$\mathrm{xy}\:+\mathrm{yz}\:+\mathrm{zx}\:>\mathrm{0}\:\:\mathrm{condition}\:\mathrm{5erified}\:\mathrm{if}\:\mathrm{a}>\mathrm{0}\:,\mathrm{b}>\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{c}>\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 09/Aug/20
$${thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 09/Aug/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$
Answered by 1549442205PVT last updated on 09/Aug/20
$$\mathrm{The}\:\mathrm{inequality}\:\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}>\sqrt{\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ $$$$\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\forall\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \mathrm{since} \\ $$$$\mathrm{squaring}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{we}\:\mathrm{get}\:\mathrm{an}\:\mathrm{equivalent}\:\mathrm{inequality} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}\sqrt{\mathrm{ab}}+\mathrm{2}\sqrt{\mathrm{bc}}+\mathrm{2}\sqrt{\mathrm{ca}}>\mathrm{a}+\mathrm{b}+\mathrm{c} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true} \\ $$
Commented by mnjuly1970 last updated on 09/Aug/20
$$\:\:\:\:\:\:\ast{thank}\:{you}\:{so}\:{much}\ast\: \\ $$$${thanks} \\ $$
Answered by udaythool last updated on 09/Aug/20
$$\mathrm{By}\:\mathrm{contradiction}\:\mathrm{method}: \\ $$$$\mathrm{If}\:\mathrm{for}\:\mathrm{some}\:{a},\:{b},\:{c}\:\in\:\mathbb{R}^{+} \\ $$$$\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\leqslant\sqrt{{a}+{b}+{c}} \\ $$$$\Rightarrow\mathrm{0}<\sqrt{{ab}}+\sqrt{{ac}}+\sqrt{{bc}}\:\leqslant\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{absurd}. \\ $$
Commented by mnjuly1970 last updated on 09/Aug/20
$${nice}\:{very}\:{nice} \\ $$
Answered by mnjuly1970 last updated on 09/Aug/20
$${anwer}:\:\mathrm{0}<\frac{{a}}{{a}+{b}+{c}}<\mathrm{1}\Rightarrow\frac{{a}}{{a}+{b}+{c}}<\sqrt{\frac{{a}}{{a}+{b}+{c}}}\:\:\ast \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\::\mathrm{0}<\frac{{b}}{{a}+{b}+{c}}<\sqrt{\frac{{b}}{{a}+{b}+{c}}}\:\:\ast\ast \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\frac{{c}}{{a}+{b}+{c}}<\sqrt{\frac{{c}}{{a}+{b}+{c}}}\:\:\ast\ast\ast \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\left(\ast\right)+\left(\ast\ast\right)+\left(\ast\ast\ast\right)\Rightarrow\mathrm{1}<\frac{\:\:\sqrt{{a}}+\sqrt{{b}}\:\:\:+\sqrt{{c}}}{\:\sqrt{{a}+{b}+{c}}} \\ $$$$\:\:\:\:\sqrt{{a}+{b}+{c}}\:<\sqrt{{a}}\:+\sqrt{{b}\:}\:+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:\:\:\:{q}.{e}.{d} \\ $$