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Question Number 211946 by BaliramKumar last updated on 25/Sep/24

Answered by BHOOPENDRA last updated on 25/Sep/24

$${(2+3+4)}^2=9^2=81$$

Commented by BHOOPENDRA last updated on 25/Sep/24

$$method\left(II\right)$$$$fromCouchyschwarzinequality$$$$(\frac{4}{x}+\frac{9}{y}+\frac{16}{z})(x+y+z)\ge {(2+3+4)}^2$$$$=81$$$$(\frac{4}{x}+\frac{9}{y}+\frac{16}{z})\ge 81$$

Answered by BHOOPENDRA last updated on 25/Sep/24

$$therearemanymethodtofindminimum$$$$valueofthat$$$$f(x,y,z)=x+y+z=1\left(1\right)$$$$g(x,y,z)=\frac{4}{x}+\frac{9}{y}+\frac{16}{z}$$$$▽f(x,y,z)=\lambda ▽g(x,y,z)$$$$▽f=(1,1,1)$$$$▽g=(-\frac{4}{x^2},\frac{-9}{y^2},\frac{-16}{z^2})$$$$(1,1,1)=(\frac{-4\lambda }{x^2},\frac{-9\lambda }{y^2},\frac{-16\lambda }{z^2})$$$$compare$$$$1=\frac{-4\lambda }{x^2},x^2=-4\lambda ,let(-\lambda )=p$$$$x=2\sqrt{p},similarly$$$$y=3\sqrt{p},z=4\sqrt{p}$$$$putthevalue$$$$x+y+z=1$$$$2\sqrt{p}+3\sqrt{p}+4\sqrt{p}=1$$$$\sqrt{p}=\frac{1}{9}$$$$p=\frac{1}{81}$$$$$$$$\frac{4}{2\times 1/9}+\frac{9}{3\times 1/9}+\frac{16}{4\times 1/9}$$$$=18+27+36=81$$

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