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Question Number 103310 by Quvonchbek last updated on 14/Jul/20

Commented by Quvonchbek last updated on 14/Jul/20

$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}$$

Answered by 1549442205 last updated on 14/Jul/20

$$\mathrm{Applying}{\mathrm{Cauchy}}^{\prime }\mathrm{s}\mathrm{inequality}\mathrm{for}$$$$\mathrm{three}\mathrm{positive}\mathrm{numbers}\mathrm{we}\mathrm{have}:$$$$\frac{{\mathrm{a}}^3}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}+\mathrm{c}}{4}+\frac{1}{2}⩾3{}^3\sqrt{\frac{{\mathrm{a}}^3}{\mathrm{b}+\mathrm{c}}.\frac{\mathrm{b}+\mathrm{c}}{4}.\frac{1}{2}}=\frac{3\mathrm{a}}{2}$$$$\mathrm{Similarly},\mathrm{we}\mathrm{have}:\frac{{\mathrm{b}}^3}{\mathrm{c}+\mathrm{a}}+\frac{\mathrm{c}+\mathrm{a}}{4}+\frac{1}{2}⩾\frac{3\mathrm{b}}{2}$$$$\frac{{\mathrm{c}}^3}{\mathrm{a}+\mathrm{b}}+\frac{\mathrm{a}+\mathrm{b}}{4}+\frac{1}{2}⩾\frac{3\mathrm{c}}{2}.\mathrm{Adding}\mathrm{three}$$$$\mathrm{above}\mathrm{inequalities}\mathrm{we}\mathrm{get}$$$$\mathrm{LHS}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}+\frac{3}{2}⩾\frac{3(\mathrm{a}+\mathrm{b}+\mathrm{c})}{2}$$$$\iff \mathrm{LHS}⩾\mathrm{a}+\mathrm{b}+\mathrm{c}-\frac{3}{2}\left(1\right)$$$$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}⩾3{}^3\sqrt{\mathrm{abc}}=3\left(2\right)(\mathrm{due}\mathrm{to}\mathrm{abc}=1$$$$\left(\mathrm{bythe}\mathrm{hypothesis}\right)).\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)$$$$\mathbf{we}\mathbf{get}\frac{{\mathbf{a}}^3}{\mathbf{b}+\mathbf{c}}+\frac{{\mathbf{b}}^3}{\mathbf{c}+\mathbf{a}}+\frac{{\mathbf{c}}^3}{\mathbf{a}+\mathbf{b}}⩾\frac{3}{2}(\mathbf{q}.\mathbf{e}.\mathbf{d})$$$$\mathbf{The}\mathbf{equality}\mathbf{ocurrs}\mathbf{if}\mathbf{and}\mathbf{only}\mathbf{if}$$$$\mathbf{a}=\mathbf{b}=\mathbf{c}=1$$$$\mathbf{second}\mathbf{way}:$$$$\mathrm{Applying}\mathrm{Cauchy}-\mathrm{Schwarz}\mathrm{we}\mathrm{have}:$$$$\frac{{\mathbf{a}}^3}{\mathbf{b}+\mathbf{c}}+\frac{{\mathbf{b}}^3}{\mathbf{c}+\mathbf{a}}+\frac{{\mathbf{c}}^3}{\mathbf{a}+\mathbf{b}}\iff \frac{{\mathbf{a}}^4}{\mathrm{a}(\mathbf{b}+\mathbf{c})}+\frac{{\mathbf{b}}^4}{\mathrm{b}(\mathbf{c}+\mathbf{a})}+\frac{{\mathbf{c}}^4}{\mathrm{c}(\mathbf{a}+\mathbf{b})}$$$$⩾\frac{{({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)}^2}{2(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}⩾\frac{{(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}^2}{2(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})}=\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{2}\left(3\right)$$$$\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{hands},$$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}⩾3{}^3\sqrt{\mathrm{ab}.\mathrm{bc}.\mathrm{ca}}=3{}^3\sqrt{{\left(\mathrm{abc}\right)}^2}=3\left(4\right)$$$$(\mathrm{due}\mathrm{to}\mathrm{abc}=1\left(\mathrm{by}\mathrm{the}\mathrm{hypothesis}\right))$$$$\mathrm{From}\left(3\right)\mathrm{and}\left(4\right)\mathrm{we}\mathrm{get}$$$$\frac{{\mathbf{a}}^3}{\mathbf{b}+\mathbf{c}}+\frac{{\mathbf{b}}^3}{\mathbf{c}+\mathbf{a}}+\frac{{\mathbf{c}}^3}{\mathbf{a}+\mathbf{b}}⩾\frac{3}{2}.\mathbf{The}\mathbf{equality}\mathbf{ocurrs}$$$$\mathbf{if}\mathbf{and}\mathbf{only}\mathbf{if}\mathbf{a}=\mathbf{b}=\mathbf{c}=1(\mathbf{q}.\mathbf{e}.\mathbf{d})$$$$$$

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