So a=(x+y)/2.
If you insert 20% into the geometric series, there is: x/b = b/c = c/y.
So xy = BCx≥b≥c≥y
(b+ 1)(c+ 1)=bc+c^2/y+b^2/x+ 1
(a+ 1)^2-(b+ 1)(c+ 1)=(x^2+y^2)/4+ 1+xy/2+x+y-bc-b-c- 1
= (x 2+y 2)/4-xy/2+x+y-b-c (xy=bc here)
=(x-y)^2/4+(x-b)+(y-c)
Because (x-y) 2 ≥ 0; (x-b)≥0； (y-c)≥0
Therefore, (a+1) 2-(b+1) (c+1) = (x-y) 2/4+(x-b)+(y-c) ≥ 0.
de:(A+ 1)2 ≥( B+ 1)(C+ 1)
24
Remember that F(x)=f(x+ 1/2),
To prove that F(x) is an even function, just prove that F(-x)=F(x),
That is, as long as it is proved that f(-x+)=f(x+ 1/2),
It is known that the images of functions f(x+ 1) and f(x) are symmetrical about y axis, and the images of functions f(x) and f(-x) are also symmetrical about y axis.
∴f(-x)=f(x+ 1).
So there is f (-x+)=f [-(x-)]
=f [(x-)+ 1]=f (x+ 1/2)
∴f(x+ 1/2) is an even function. /