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Question Number 107242 by mnjuly1970 last updated on 09/Aug/20

...question... prove that: if a,b,c∈R^+ then: ♣ (√a) +(√b)+(√c)> (√(a+b+c)) ♣ ....sincerly yours... ... M.N...

$$\:\:\:\:\:...{question}... \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:{that}: \\ $$ $$\:{if}\:\:{a},{b},{c}\in\mathbb{R}^{+} \:{then}: \\ $$ $$\:\:\:\:\:\clubsuit\:\:\:\sqrt{{a}}\:+\sqrt{{b}}+\sqrt{{c}}>\:\sqrt{{a}+{b}+{c}}\:\clubsuit\: \\ $$ $$\:\:\:\:\:\:\:....{sincerly}\:{yours}... \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\mathscr{M}.\mathscr{N}... \\ $$ $$\:\: \\ $$

Answered by abdomathmax last updated on 09/Aug/20

a= x^2 ,b=y^2 ,c =z^2 (i)⇒x+y +z>(√(x^2 +y^2 +z^2 )) ⇒(x+y+z)^2 >x^2 +y^2 +z^2 ⇒ x^2 +y^2 +z^(2 ) +2(xy +yz +zx)>x^2 +y^2 +z^2 xy +yz +zx >0 condition 5erified if a>0 ,b>0 and c>0

$$\mathrm{a}=\:\mathrm{x}^{\mathrm{2}} \:,\mathrm{b}=\mathrm{y}^{\mathrm{2}} \:,\mathrm{c}\:=\mathrm{z}^{\mathrm{2}} \:\:\:\:\left(\mathrm{i}\right)\Rightarrow\mathrm{x}+\mathrm{y}\:+\mathrm{z}>\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}} } \\ $$ $$\Rightarrow\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} >\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}} \:\Rightarrow \\ $$ $$\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}\:} \:+\mathrm{2}\left(\mathrm{xy}\:+\mathrm{yz}\:+\mathrm{zx}\right)>\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{2}} \\ $$ $$\mathrm{xy}\:+\mathrm{yz}\:+\mathrm{zx}\:>\mathrm{0}\:\:\mathrm{condition}\:\mathrm{5erified}\:\mathrm{if}\:\mathrm{a}>\mathrm{0}\:,\mathrm{b}>\mathrm{0} \\ $$ $$\mathrm{and}\:\mathrm{c}>\mathrm{0} \\ $$

Commented bymnjuly1970 last updated on 09/Aug/20

thank you

$${thank}\:{you} \\ $$

Commented bymathmax by abdo last updated on 09/Aug/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Answered by 1549442205PVT last updated on 09/Aug/20

The inequality (√a)+(√b)+(√c)>(√(a+b+c)) is always true ∀a,b,c∈R^+ since squaring two sides we get an equivalent inequality a+b+c+2(√(ab))+2(√(bc))+2(√(ca))>a+b+c which is always true

$$\mathrm{The}\:\mathrm{inequality}\:\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{c}}>\sqrt{\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ $$ $$\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\forall\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \mathrm{since} \\ $$ $$\mathrm{squaring}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{we}\:\mathrm{get}\:\mathrm{an}\:\mathrm{equivalent}\:\mathrm{inequality} \\ $$ $$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}\sqrt{\mathrm{ab}}+\mathrm{2}\sqrt{\mathrm{bc}}+\mathrm{2}\sqrt{\mathrm{ca}}>\mathrm{a}+\mathrm{b}+\mathrm{c} \\ $$ $$\mathrm{which}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true} \\ $$

Commented bymnjuly1970 last updated on 09/Aug/20

∗thank you so much∗ thanks

$$\:\:\:\:\:\:\ast{thank}\:{you}\:{so}\:{much}\ast\: \\ $$ $${thanks} \\ $$

Answered by udaythool last updated on 09/Aug/20

By contradiction method: If for some a, b, c ∈ R^+ (√a)+(√b)+(√c)≤(√(a+b+c)) ⇒0<(√(ab))+(√(ac))+(√(bc)) ≤0 which is absurd.

$$\mathrm{By}\:\mathrm{contradiction}\:\mathrm{method}: \\ $$ $$\mathrm{If}\:\mathrm{for}\:\mathrm{some}\:{a},\:{b},\:{c}\:\in\:\mathbb{R}^{+} \\ $$ $$\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\leqslant\sqrt{{a}+{b}+{c}} \\ $$ $$\Rightarrow\mathrm{0}<\sqrt{{ab}}+\sqrt{{ac}}+\sqrt{{bc}}\:\leqslant\mathrm{0} \\ $$ $$\mathrm{which}\:\mathrm{is}\:\mathrm{absurd}. \\ $$

Commented bymnjuly1970 last updated on 09/Aug/20

nice very nice

$${nice}\:{very}\:{nice} \\ $$

Answered by mnjuly1970 last updated on 09/Aug/20

anwer: 0<(a/(a+b+c))<1⇒(a/(a+b+c))<(√(a/(a+b+c))) ∗ :0<(b/(a+b+c))<(√(b/(a+b+c))) ∗∗ 0<(c/(a+b+c))<(√(c/(a+b+c))) ∗∗∗ ∴ (∗)+(∗∗)+(∗∗∗)⇒1<(( (√a)+(√b) +(√c))/(√(a+b+c))) (√(a+b+c)) <(√a) +(√(b )) +(√c) q.e.d

$${anwer}:\:\mathrm{0}<\frac{{a}}{{a}+{b}+{c}}<\mathrm{1}\Rightarrow\frac{{a}}{{a}+{b}+{c}}<\sqrt{\frac{{a}}{{a}+{b}+{c}}}\:\:\ast \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\::\mathrm{0}<\frac{{b}}{{a}+{b}+{c}}<\sqrt{\frac{{b}}{{a}+{b}+{c}}}\:\:\ast\ast \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\frac{{c}}{{a}+{b}+{c}}<\sqrt{\frac{{c}}{{a}+{b}+{c}}}\:\:\ast\ast\ast \\ $$ $$\therefore\:\:\:\:\:\:\:\:\:\left(\ast\right)+\left(\ast\ast\right)+\left(\ast\ast\ast\right)\Rightarrow\mathrm{1}<\frac{\:\:\sqrt{{a}}+\sqrt{{b}}\:\:\:+\sqrt{{c}}}{\sqrt{{a}+{b}+{c}}} \\ $$ $$\:\:\:\:\sqrt{{a}+{b}+{c}}\:<\sqrt{{a}}\:+\sqrt{{b}\:}\:+\sqrt{{c}} \\ $$ $$\:\:\:\:\:\:\:\:\:{q}.{e}.{d} \\ $$

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