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

$$...question...$$ $$provethat:$$ $$ifa,b,c\in {\mathbb{R}}^{+}then:$$ $$\mathrm{♣}\sqrt{a}+\sqrt{b}+\sqrt{c}>\sqrt{a+b+c}\mathrm{♣}$$ $$....sincerlyyours...$$ $$...\mathcal{M}.\mathcal{N}...$$ $$$$

Answered by abdomathmax last updated on 09/Aug/20

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

Commented bymnjuly1970 last updated on 09/Aug/20

$$thankyou$$

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

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

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

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

Commented bymnjuly1970 last updated on 09/Aug/20

$$\ast thankyousomuch\ast$$ $$thanks$$

Answered by udaythool last updated on 09/Aug/20

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

Commented bymnjuly1970 last updated on 09/Aug/20

$$niceverynice$$

Answered by mnjuly1970 last updated on 09/Aug/20

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

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