If x1,x2,⋯xn are n non-zero real numbers such that x12+x22+⋯+xn−12x22+x32+⋯+xn2≤x1x2+x2x3+⋯+xn−1xn2 then x1,x2,⋯,xnare in

We shall make use of the identitya12+a22+⋯+am2b12+b22+⋯+bm2−a1b1+a2b2+⋯+ambm2=a1b2−a2b12+a1b3−a3b12+⋯+am−1bm−ambm−12Thus, x12+x22+⋯+xn−12x22+x32+⋯+xn2−x1x2+x2x3+⋯+xn−1xn2≤0⇒x1x3−x2x22+x1x4−x3x32+⋯+xn−2xn−xn−1xn−12≤0As x1,x2,⋯,xn are real, this is possible if and only ifx1x3−x22=x2x4−x32=⋯=xn−2xn−xn−12=0⇒x1x2=x2x3=x3x4=⋯=xnxn−1⇒x1,x2,⋯,xn are in G.P.

If x1,x2,⋯xn are n non-zero real numbers such that x12+x22+⋯+xn−12x22+x32+⋯+xn2≤x1x2+x2x3+⋯+xn−1xn2 then x1,x2,⋯,xnare in

We shall make use of the identitya12+a22+⋯+am2b12+b22+⋯+bm2−a1b1+a2b2+⋯+ambm2=a1b2−a2b12+a1b3−a3b12+⋯+am−1bm−ambm−12Thus, x12+x22+⋯+xn−12x22+x32+⋯+xn2−x1x2+x2x3+⋯+xn−1xn2≤0⇒x1x3−x2x22+x1x4−x3x32+⋯+xn−2xn−xn−1xn−12≤0As x1,x2,⋯,xn are real, this is possible if and only ifx1x3−x22=x2x4−x32=⋯=xn−2xn−xn−12=0⇒x1x2=x2x3=x3x4=⋯=xnxn−1⇒x1,x2,⋯,xn are in G.P.

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If $x_{1}, x_{2}, \dots x_{n}$ are n non-zero real numbers such that $(x_{1}^{2} + x_{2}^{2} + \dots + x_{n - 1}^{2}) (x_{2}^{2} + x_{3}^{2} + \dots + x_{n}^{2}) \leq$ ${(x_{1} x_{2} + x_{2} x_{3} + \dots + x_{n - 1} x_{n})}^{2}$ then $x_{1}, x_{2}, \dots, x_{n}$
are in

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Detailed Solution

We shall make use of the identity

$\begin{array}{r} (a_{1}^{2} + a_{2}^{2} + \dots + a_{m}^{2}) (b_{1}^{2} + b_{2}^{2} + \dots + b_{m}^{2}) - \\ {(a_{1} b_{1} + a_{2} b_{2} + \dots + a_{m} b_{m})}^{2} \\ = {(a_{1} b_{2} - a_{2} b_{1})}^{2} + {(a_{1} b_{3} - a_{3} b_{1})}^{2} + \dots + \\ {(a_{m - 1} b_{m} - a_{m} b_{m - 1})}^{2} \end{array}$

Thus, $\begin{aligned} (x_{1}^{2} + x_{2}^{2} + \dots + x_{n - 1}^{2}) (x_{2}^{2} + x_{3}^{2} + \dots + x_{n}^{2}) \\ - {(x_{1} x_{2} + x_{2} x_{3} + \dots + x_{n - 1} x_{n})}^{2} \leq 0 \\ \Rightarrow {(x_{1} x_{3} - x_{2} x_{2})}^{2} + {(x_{1} x_{4} - x_{3} x_{3})}^{2} + \dots \\ + {(x_{n - 2} x_{n} - x_{n - 1} x_{n - 1})}^{2} \leq 0 \end{aligned}$

As $x_{1}, x_{2}, \dots, x_{n}$ are real, this is possible if and only if

$\begin{array}{l} x_{1} x_{3} - x_{2}^{2} = x_{2} x_{4} - x_{3}^{2} = \dots = x_{n - 2} x_{n} - x_{n - 1}^{2} = 0 \\ \Rightarrow \frac{x_{1}}{x_{2}} = \frac{x_{2}}{x_{3}} = \frac{x_{3}}{x_{4}} = \dots = \frac{x_{n}}{x_{n - 1}} \end{array}$