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Question Number 98570 by HamraboyevFarruxjon last updated on 14/Jun/20

$$\mathit{a},\mathit{b},\mathit{c}>0\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}:$$ $$\frac{\mathit{a}}{\sqrt{{\mathit{a}}^2+8\mathit{b}\mathit{c}}}+\frac{\mathit{b}}{\sqrt{{\mathit{b}}^2+8\mathit{a}\mathit{c}}}+\frac{\mathit{c}}{\sqrt{{\mathit{c}}^2+8\mathit{a}\mathit{b}}}⩾1$$ $$\mathit{h}\mathit{e}\mathit{l}\mathit{p}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}...$$

Commented byMJS last updated on 14/Jun/20

$$\mathrm{extremes}\mathrm{where}a=b=c$$ $$\mathrm{here}\mathrm{this}\mathrm{leads}\mathrm{to}$$ $$3\frac{a}{\sqrt{a^2+8a^2}}=1\mathrm{for}a>0$$ $$\Rightarrow \min \left(\mathrm{lhs}\right)=1\Rightarrow \mathrm{proven}$$

Answered by 1549442205 last updated on 15/Jun/20

$$$$ $$$$ $$\mathbf{Applying}\mathbf{the}\mathbf{Cauchi}-\mathbf{Schwartz}\mathbf{we}\mathbf{have}$$ $$\mathrm{we}\mathrm{have}\mathrm{P}=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+8\mathrm{bc}}}+\frac{\mathbf{b}}{\sqrt{{\mathbf{b}}^2+8\mathbf{ca}}}+\frac{\mathbf{c}}{\sqrt{{\mathbf{c}}^2+8\mathbf{ab}}}=\frac{{\mathbf{a}}^2}{\mathbf{a}\sqrt{{\mathbf{a}}^2+8\mathbf{bc}}}+\frac{{\mathbf{b}}^2}{\mathbf{b}\sqrt{{\mathbf{b}}^2+8\mathbf{ca}}}+\frac{{\mathbf{c}}^2}{\mathbf{c}\sqrt{{\mathbf{c}}^2+8\mathbf{ab}}}⩾\frac{{(\mathbf{a}+\mathbf{b}+\mathbf{c})}^2}{\mathbf{a}\sqrt{{\mathbf{a}}^2+8\mathbf{bc}}+\mathbf{b}\sqrt{{\mathbf{b}}^2+8\mathbf{ca}}+\mathbf{c}\sqrt{{\mathbf{c}}^2+8\mathbf{ab}}}$$ $$\mathbf{On}\mathbf{the}\mathbf{other}\mathbf{hands},\mathbf{also}\mathbf{by}\mathbf{C}-\mathbf{S}\mathbf{we}\mathbf{have}$$ $$\mathbf{a}\sqrt{{\mathbf{a}}^2+8\mathbf{bc}}+\mathbf{b}\sqrt{{\mathbf{b}}^2+8\mathbf{ca}}+\mathbf{c}\sqrt{{\mathbf{c}}^2+8\mathbf{ab}}=\sqrt{\mathbf{a}}.\sqrt{{\mathbf{a}}^3+8\mathbf{abc}}+\sqrt{\mathbf{b}}.\sqrt{{\mathbf{b}}^3+8\mathbf{abc}}+\sqrt{\mathbf{c}}.\sqrt{{\mathbf{c}}^3+8\mathbf{abc}}$$ $$⩽\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})({\mathbf{a}}^3+{\mathbf{b}}^3+{\mathbf{c}}^3+24\mathbf{abc})}.\mathbf{Hence},$$ $$\mathbf{P}⩾\frac{{(\mathbf{a}+\mathbf{b}+\mathbf{c})}^2}{\sqrt{(\mathbf{a}+\mathbf{b}+\mathbf{c})({\mathbf{a}}^3+{\mathbf{b}}^3+{\mathbf{c}}^3+24\mathbf{abc})}}=\sqrt{\frac{{(\mathbf{a}+\mathbf{b}+\mathbf{c})}^3}{{\mathbf{a}}^3+{\mathbf{b}}^3+{\mathbf{c}}^3+24\mathbf{abc}}}.\mathbf{Therefore},\mathbf{it}\mathbf{is}\mathbf{enough}$$ $$\mathbf{to}\mathbf{prove}\mathbf{that}\frac{{(\mathbf{a}+\mathbf{b}+\mathbf{c})}^3}{{\mathbf{a}}^3+{\mathbf{b}}^3+{\mathbf{c}}^3+24\mathbf{abc}}⩾1\iff$$ $${(\mathbf{a}+\mathbf{b}+\mathbf{c})}^3⩾{\mathbf{a}}^3+{\mathbf{b}}^3+{\mathbf{c}}^3+24\mathbf{abc}\iff (\mathbf{a}+\mathbf{b})(\mathbf{b}+\mathbf{c})(\mathbf{c}+\mathbf{a})⩾8\mathbf{abc}$$ $$\mathbf{This}\mathbf{final}\mathbf{inequality}\mathbf{is}\mathbf{always}\mathbf{true}\mathbf{because}$$ $$\mathbf{it}\mathbf{is}\mathbf{followed}\mathbf{from}\mathbf{the}\mathbf{inequlities}:$$ $$\mathbf{a}+\mathbf{b}⩾2\sqrt{\mathbf{ab}},\mathbf{b}+\mathbf{c}⩾2\sqrt{\mathbf{bc}},\mathbf{c}+\mathbf{a}⩾2\sqrt{\mathbf{ca}}$$ $$\mathbf{Thus},\mathbf{P}⩾1.\mathbf{The}\mathbf{equality}\mathbf{occurs}\mathbf{if}\mathbf{an}\mathbf{only}\mathbf{if}$$ $$\mathbf{a}=\mathbf{b}=\mathbf{c}(\mathbf{q}.\mathbf{e}.\mathbf{d})$$

Commented byFarruxjano last updated on 15/Jun/20

$$\mathrm{Sir}\mathrm{thanks}\mathrm{a}\mathrm{lot}$$

Commented by1549442205 last updated on 25/Jun/20

$$\mathrm{Thank}\mathrm{you}!\mathrm{you}\mathrm{are}\mathrm{wellcome},\mathrm{sir}.$$

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