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Question Number 49604 by Tawa1 last updated on 08/Dec/18

$$\mathrm{Show}\mathrm{that}:\frac{{(\mathrm{a}+\mathrm{b})}^2}{2}⩽{\mathrm{a}}^2+{\mathrm{b}}^2$$

Answered by afachri last updated on 08/Dec/18

$$\mathrm{let}{(a-b)}^2⩾0$$$$a^2-2ab+b^2⩾0$$$$a^2+b^2⩾2ab$$$$a^2+2ab+b^2⩾2ab+2ab$$$${(a+b)}^2⩾4\left(ab\right)$$$$$$$$\mathrm{meanwhile}:ab=\frac{{(a+b)}^2-(a^2+b^2)}{2}$$$${(a+b)}^2⩾4\left(\frac{{(a+b)}^2-(a^2+b^2)}{2}\right)$$$${(a+b)}^2⩾2{(a+b)}^2-2(a^2+b^2)]$$$$-{(a+b)}^2⩾-2(a^2+b^2)$$$$\frac{{(a+b)}^2}{2}⩽(a^2+b^2)$$$$$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{Sir}\mathrm{please}\mathrm{help}\mathrm{with}\mathrm{this}\mathrm{too}$$$$$$$$\mathrm{Show}\mathrm{that}:\sqrt[4]{\mathrm{abcd}}=\frac{1}{4}(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})$$

Commented by afachri last updated on 08/Dec/18

$$\mathrm{Arithmaric}\mathrm{Mean}=\frac{a+b+c+d}{4}$$$$\mathrm{Geometric}\mathrm{Mean}=\sqrt[4]{{abcd}^{}}$$$$$$$$\mathrm{Arithmatic}\mathrm{Mean}⩾\mathrm{Geometric}\mathrm{Mean}$$$$\frac{1}{4}(a+b+c+d)⩾\sqrt[4]{{abcd}^{}}$$$$\mathrm{so},\mathrm{the}\mathrm{equalty}\mathrm{can}\mathrm{be}\mathrm{achieved}\mathrm{only}$$$$\mathrm{and}\mathrm{if}\mathrm{only}:$$$$a=b=c=d$$$$\mathrm{then}$$$$\frac{1}{4}(a+b+c+d)=\sqrt[4]{{abcd}^{}}$$$$$$$$$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by afachri last updated on 08/Dec/18

$$\mathrm{ur}\mathrm{welcome},\mathrm{Sir}.$$$$\mathrm{Pardon}\mathrm{me},\mathrm{are}\mathrm{u}\mathrm{an}\mathrm{Indonesian}?$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{No}\mathrm{sir}$$

Commented by afachri last updated on 08/Dec/18

$$\mathrm{nevermind}\mathrm{Sir}.$$$${\mathrm{i}}^{\prime }\mathrm{m}\mathrm{just}\mathrm{asking}\mathrm{Sir}.:)$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{Ok}\mathrm{sir}$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{What}\mathrm{if}\mathrm{the}\mathrm{first}\mathrm{question}\mathrm{is}:{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}^2⩽\frac{\mathrm{a}+\mathrm{b}}{2}$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{How}\mathrm{will}\mathrm{the}\mathrm{prove}\mathrm{be}$$

Commented by afachri last updated on 08/Dec/18

$$\mathrm{i}\mathrm{had}\mathrm{given}\mathrm{the}2\mathrm{solutuions}\mathrm{earlier}\mathrm{Sir}.$$$$\mathrm{what}\mathrm{kind}\mathrm{of}\mathrm{proof}\mathrm{else}\mathrm{you}\mathrm{sesrching}$$$$\mathrm{for}\mathrm{Sir}??{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{sorry}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{get}\mathrm{it}.$$$$$$

Answered by afachri last updated on 08/Dec/18

$$\mathrm{according}\mathrm{to}\mathbf{QM}-\mathbf{AM}$$$$\mathbf{QM}⩾\mathbf{AM}$$$$\sqrt{\frac{a^{2^{}}+b^2}{2}}⩾\frac{a+b}{2}\mathbf{then}\mathbf{square}\mathbf{both}\mathbf{sides}.$$$$\frac{a^2+b^2}{2}⩾\frac{{(a+b)}^2}{4_{}}$$$$a^2+b^2⩾\frac{{(a+b)}^2}{2}$$$$$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{What}\mathrm{is}\mathrm{QM}\mathrm{and}\mathrm{AM}\mathrm{sir}$$

Commented by afachri last updated on 08/Dec/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome},\mathrm{Sir}$$

Commented by Tawa1 last updated on 08/Dec/18

$$???$$

Commented by afachri last updated on 08/Dec/18

$$\mathrm{quadratic}\mathrm{mean}\mathrm{and}$$$$\mathrm{arithmaric}\mathrm{mean}$$

Commented by Tawa1 last updated on 08/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$$$$$

Commented by afachri last updated on 08/Dec/18

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{been}\mathrm{my}\mathrm{pleasure}\mathrm{Sir}$$

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