squeeze theorem of lim(x,y,z) (x^2y^2z^2)/(x^2+y^2+z^2)

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We know that Arithmetic Mean (AM) is greater or equal to Geometric mean(GM). The statement of A M \geq G M is, if x_{1}, x_{2}, \ldots x_{n} be positive real number then \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \geq \sqrt[n]{x_{1} \cdot x_{2} \cdots \cdots x_{n}} equality occurs if x_{1}=x_{2}=\cdots=x_{n}Here we have to find \lim _{(x, y, z) \rightarrow(0,0,0)} \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}}.Now we know that x^{2}, y^{2}, z^{2} all are positive. Now apply A M \geq G M on x^{2}, y^{2}, z^{2} then we get, \frac{x^{2}+y^{2}+z^{2}}{3} \geq \sqrt[3]{x^{2} y^{2} z^{2}}Now \frac{x^{2}+y^{2}+z^{2}}{3} \geq \sqrt[3]{x^{2} y^{2} z^{2}} \Rightarrow \frac{x^{2}+y^{2}+z^{2}}{3} \geq(x y z)^{\frac{2}{3}} \Rightarrow \frac{1}{x^{2}+y^{2}+z^{2}} \leq \frac{3}{(x y z)^{\frac{2}{3}}} \Rightarrow \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}} \leq \frac{3(x y z)^{2}}{(x y z)^{\frac{2}{3}}}\Rightarrow \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}} \leq 3(x y z)^{2-\frac{2}{3}} \Rightarrow \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}} \leq 3(x y z)^{\frac{4}{3}}Now \lim _{(x, y, z) \rightarrow(0,0,0)} 3(x y z)^{\frac{4}{3}}=3 \lim _{(x, y, z) \rightarrow(0,0,0)}(x y z)^{\frac{4}{3}}=3 \times 0=0We also have x^{2} \geq 0 for any real x then \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}} \geq 0.Now then we have, 0 \leq \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}} \leq 3(x y z)^{\frac{4}{3}}.Now since \lim _{(x, y, z) \rightarrow(0,0,0)} 3(x y z)^{\frac{4}{3}}=0 then by Squeeze theorem we can conclude that \lim _{x, y, z) \rightarrow(0,0,0)} \frac{x^{2} y^{2} z^{2}}{x^{2}+y^{2}+z^{2}}=0 Happy ing :) ...