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Question

Let $0 < z < y < x$ be three real numbers such that $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in an arithmetic progression and $$x, \sqrt{2} y, z$$ are in a geometric progression. If $$x y+y z+z x=\frac{3}{\sqrt{2}} x y z$$

Accepted Answer

$\because \frac{1}{x},\frac{1}{y},\frac{1}{z}$ are in A.P.

$\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y}$ ........... (i)

and $x,\sqrt{2}y,z$ are in G.P.

$\Rightarrow 2{y}^{2}=xz$ .......... (ii)

from (i), $\frac{2}{y}=\frac{x+z}{xz}=\frac{x+z}{2{y}^{2}}$

$\Rightarrow 4y=x+z$

$\begin{gathered} xy+yz+zx=\frac{3}{\sqrt{2}}xyz \ y(4y)+xz=\frac{3}{\sqrt{2}}(2{y}^{2})y \ \Rightarrow 4{y}^{2}+2{y}^{2}=3\sqrt{2}{y}^{3} \ \Rightarrow 6{y}^{2}=3\sqrt{2}{y}^{3}\Rightarrow y=\sqrt{2} \ 	herefore 3(x+y+z{)}^{2}=3(5y{)}^{2}=3(5\sqrt{2}{)}^{2} \ =150 \end{gathered}$

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