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 74213 by FCB last updated on 20/Nov/19

Commented by MJS last updated on 20/Nov/19

information missing I think the question is find minimum of the expression this should be at a=b=c=(1/3) in this case the answer is ((69)/4)=17.25

$$\mathrm{information}\:\mathrm{missing} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{find}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{expression} \\ $$$$\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{at}\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{69}}{\mathrm{4}}=\mathrm{17}.\mathrm{25} \\ $$

Answered by mind is power last updated on 20/Nov/19

what is quation ?

$${what}\:{is}\:{quation}\:? \\ $$

Answered by mind is power last updated on 20/Nov/19

a=b=0,c=1 =14+(7/2)=((35)/2)=17,5 min((7+2b)/(1+a))+((7+2c)/(1+b))+((7+2a)/(1+c)) a+b+c=1,a,b,c≥0 x=1+a,y=1+b,z=1+c⇒x+y+z=4 ⇔f(x,y,z)((2y+5)/x)+((5+2z)/y)+((5+2x)/z)=5((1/x)+(1/y)+(1/z))+2((y/x)+(z/y)+(x/z)) we will use Ha−GM−AR inqualite (3/((1/x)+(1/y)+(1/z)))≤((xyz))^(1/3) ≤((x+y+z)/3)⇒(1/x)+(1/y)+(1/z)≥(9/(x+y+z))=(9/4) ((y/x)+(z/y)+(x/z))≥3(((((y.x.z)/(x.y.z)))))^(1/3) =3 ⇒f(x,y,z)≥6+(9/4).5=6+11.25=17.25 equalitiy when x=y=z=(4/3)⇔a=b=c=(1/3)

$${a}={b}=\mathrm{0},{c}=\mathrm{1} \\ $$$$=\mathrm{14}+\frac{\mathrm{7}}{\mathrm{2}}=\frac{\mathrm{35}}{\mathrm{2}}=\mathrm{17},\mathrm{5}\: \\ $$$${min}\frac{\mathrm{7}+\mathrm{2}{b}}{\mathrm{1}+{a}}+\frac{\mathrm{7}+\mathrm{2}{c}}{\mathrm{1}+{b}}+\frac{\mathrm{7}+\mathrm{2}{a}}{\mathrm{1}+{c}} \\ $$$${a}+{b}+{c}=\mathrm{1},{a},{b},{c}\geqslant\mathrm{0} \\ $$$${x}=\mathrm{1}+{a},{y}=\mathrm{1}+{b},{z}=\mathrm{1}+{c}\Rightarrow{x}+{y}+{z}=\mathrm{4} \\ $$$$\Leftrightarrow{f}\left({x},{y},{z}\right)\frac{\mathrm{2}{y}+\mathrm{5}}{{x}}+\frac{\mathrm{5}+\mathrm{2}{z}}{{y}}+\frac{\mathrm{5}+\mathrm{2}{x}}{{z}}=\mathrm{5}\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)+\mathrm{2}\left(\frac{{y}}{{x}}+\frac{{z}}{{y}}+\frac{{x}}{{z}}\right) \\ $$$${we}\:{will}\:{use}\:{Ha}−{GM}−{AR}\:{inqualite} \\ $$$$\frac{\mathrm{3}}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}}\leqslant\sqrt[{\mathrm{3}}]{{xyz}}\leqslant\frac{{x}+{y}+{z}}{\mathrm{3}}\Rightarrow\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\geqslant\frac{\mathrm{9}}{{x}+{y}+{z}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\left(\frac{{y}}{{x}}+\frac{{z}}{{y}}+\frac{{x}}{{z}}\right)\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\left(\frac{{y}.{x}.{z}}{{x}.{y}.{z}}\right)}=\mathrm{3} \\ $$$$\Rightarrow{f}\left({x},{y},{z}\right)\geqslant\mathrm{6}+\frac{\mathrm{9}}{\mathrm{4}}.\mathrm{5}=\mathrm{6}+\mathrm{11}.\mathrm{25}=\mathrm{17}.\mathrm{25}\: \\ $$$${equalitiy}\:{when}\:{x}={y}={z}=\frac{\mathrm{4}}{\mathrm{3}}\Leftrightarrow{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com