Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 13153 by Tinkutara last updated on 15/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

$$(1+a)(1+b)(1+c)(1+d)=$$$$(abcd+a)(abcd+b)(abcd+c)(abcd+d)=$$$$abcd(dbc+1)(acd+1)(abd+1)(abc+1)=$$$$(abc+1)(abd+1)(acd+1)(dbc+1)⩾$$$$⩾2\sqrt{abc\times 1}\times 2\sqrt{abd\times 1}\times 2\sqrt{acd\times 1}\times 2\sqrt{dbc\times 1}=$$$$⩾16abcd\sqrt{abcd}=16.◼$$

Commented by Tinkutara last updated on 16/May/17

$$\mathrm{Thanks}\mathrm{behi}!$$

Commented by RasheedSindhi last updated on 20/May/17

$$\mathcal{V}\mathcal{N}ice!!!$$

Answered by mrW1 last updated on 16/May/17

$$since\frac{x+y}{2}⩾\sqrt{xy},wehave$$$$1+a⩾2\sqrt{a}$$$$1+b⩾2\sqrt{b}$$$$1+c⩾2\sqrt{c}$$$$1+d⩾2\sqrt{d}$$$$\Rightarrow (1+a)(1+b)(1+c)(1+d)⩾2^4\sqrt{abcd}=16\sqrt{1}=16$$

Commented by Tinkutara last updated on 16/May/17

$$\mathrm{Thanks}\mathrm{mrW}1!$$

Commented by RasheedSindhi last updated on 20/May/17

$$\mathcal{N}ice\mathrm{\&}\mathcal{S}mart!!!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com