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Mathematical inequality in senior high school
(1) The slope of the straight line L is 2xn, passing through (xn, yn) (xn+ 1, 0).

And because (xn, yn) is on the curve, yn = xn 2+2.

By the slope formula k=(y 1-y2)/(x 1-x2)

get 2xn =(xn ^ 2+2)/(xn-xn+ 1)

Simplified to xn+1= (xn 2+2)/(2xn)

(2) suppose Xn+ 1-√ 2.

Get xn & gt√2

That is, as long as xn >; is proved; √2, the original inequality holds.

Prove:

xn+ 1=(xn^2+2)/(2xn)=(xn/2)+( 1/xn)

Because x1= 2 >; 0, xn is always greater than 0.

Xn+ 1=(xn/2)+(2/xn)>=2√(xn/2)*( 1/xn)=√2

That is, xn & gt√2 proposition proof

(3)xn+ 1-xn =(xn/2)+( 1/xn)-xn =(2-xn 2)/2xn

Because xn > 2, xn 2 > 2.

That is, xn+ 1-xn < 0, that is, xn monotonically decreases.

Xn is always less than or equal to 2 because x 1=2.

That is, the range of a is (2, positive infinity)