Question

If $x,y,z\in +R$ satisfies $x+y+z=1$, then the minimum value of $(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})$, is

• A. $8$
• B. $16$
• C. $32$
• D. $64$

Answer 1

The correct option is D $64$

Given, $x+y+z=1$

we know that, $A.M.\ge G.M.$

$⟹\frac{x+y+z}{3}\ge \sqrt[3]{3xyz}$

$⟹\frac{1}{3}\ge \sqrt[3]{3xyz}$

$⟹\sqrt[3]{3xyz}\le \frac{1}{3}$

cubing on both sides

$⟹xyz\le \frac{1}{27}$ --------------(1)

Similarly

$⟹\frac{1+x+1+y+1+z}{3}\ge \sqrt[3]{3(1+x)(1+y)(1+z)}$

$⟹\frac{3+(x+y+z)}{3}\ge \sqrt[3]{3(1+x)(1+y)(1+z)}$

$⟹\frac{3+1}{3}\ge \sqrt[3]{3(1+x)(1+y)(1+z)}$

$⟹\frac{4}{3}\ge \sqrt[3]{3(1+x)(1+y)(1+z)}$

cubing on both sides

$⟹\frac{1}{(1+x)(1+y)(1+z)}\le \frac{27}{64}$ ---------------(2)

multiplying (1) and (2) we get

$⟹\frac{xyz}{(1+x)(1+y)(1+z)}\le \frac{1}{27}\ast \frac{27}{64}$

$⟹\frac{xyz}{(1+x)(1+y)(1+z)}\le \frac{1}{64}$

$⟹\frac{1}{(1+x)(1+y)(1+z)}\le \frac{1}{64xyz}$

$⟹\frac{(1+x)(1+y)(1+z)}{xyz}\ge 64$

$⟹\frac{(1+x)}{x}\ast \frac{(1+y)}{y}\ast \frac{(1+z)}{z}\ge 64$

$⟹(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})\ge 64$

Therefore, the minimum value of $(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})$ is $64$