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Question Number 55150 by naka3546 last updated on 18/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19

$$\frac{(1+x)+(1+y)+(1+z)}{3}⩾\sqrt[3]{(1+x)(1+y)(1+z)}⩾\frac{3}{\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}}$$$$1+\frac{x+y+z}{3}⩾\frac{3}{2}$$$$\frac{x+y+z}{3}⩾\frac{1}{2}$$$$x+y+z⩾\frac{3}{2}$$$$now$$$$\frac{1+4x+1+4y+1+4z}{3}⩾\frac{3}{\frac{1}{1+4x}+\frac{1}{1+4y}+\frac{1}{1+4z}}$$$$1+\frac{4}{3}(x+y+z)⩾\frac{3}{k}$$$$1+\frac{4}{3}\times \frac{3}{2}⩾\frac{3}{k}$$$$3⩾\frac{3}{k}$$$$3k⩾3$$$$sominimumvalueof\frac{1}{1+4x}+\frac{1}{1+4y}+\frac{1}{1+4z}=1$$

Answered by tm888 last updated on 19/Feb/19

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