Question

Square root of $\frac{1}{xyz}(x^2+y^2+z^2)+2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})is$

• A. $\sqrt{\frac{x}{yz}}+\sqrt{\frac{y}{zx}}+\sqrt{\frac{z}{xy}}$
• B. $\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}$
• C. ${\left(\frac{x}{yz}\right)}^2+{\left(\frac{y}{zx}\right)}^2+{\left(\frac{z}{xy}\right)}^2$
• D. ${\left(\frac{yz}{x}\right)}^{\frac{1}{2}}+{\left(\frac{zx}{y}\right)}^{\frac{1}{2}}+{\left(\frac{xy}{z}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is A $\sqrt{\frac{x}{yz}}+\sqrt{\frac{y}{zx}}+\sqrt{\frac{z}{xy}}$

we have,

formula,

$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

Given,

$\frac{1}{xyz}(x^2+y^2+z^2)+2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

$=\frac{x}{yz}+\frac{y}{xz}+\frac{z}{yx}+2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

comparing the above equation with formula, we get,

$a^2=\frac{x}{yz}$

$\Rightarrow a=\sqrt{\frac{x}{yz}}$

similarly

$b^2=\frac{y}{xz}$

$\Rightarrow b=\sqrt{\frac{y}{xz}}$

and

$c^2=\frac{z}{xy}$

$\Rightarrow c=\sqrt{\frac{z}{yx}}$

$\therefore {(\sqrt{\frac{x}{yz}}+\sqrt{\frac{y}{xz}}+\sqrt{\frac{z}{yx}})}^2=\frac{1}{xyz}(x^2+y^2+z^2)+2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$