It is known that a, b, x, y are positive real numbers, and a²+b²=1, x²+y²=1.

It is known that a, b, x, y are positive real numbers, and a²+b²=1, x²+y²=1.

Proof: ax+by≤1

parsing: Because a, b, x, and y are all positive real numbers,

So according to a²+b²=1, x²+y²=1,

can Construct two right triangles,

, but because the two right triangles are equal in oblique sides,

We can think of the circumference angle of the circle to which the diameter of the circle is 90°,

Constructs two right triangles to which the diameter of a circle is opposite,

as shown in the figure below:

in ⊙O, diameter AB=1,

in Rt△ACB,

Let AC=a, BC=b

from a²+b²=AC²+CB²=AB²=1;

In Rt△ADB,

Let BD=x, AD=y,

from x²+y²=BD²+AD²=AB²=1;

In summary, the constructed ⊙O and two right triangles are in line with the question,

then ax+by=AC·BD+CB·AD.

If you still remember the Ptolemy theorem,

then there is:

AC·BD+CB·AD=AB·CD.

Because AB is the diameter,

so CD≤AB,

so AC·BD+CB·AD≤AB²=1,

that is ax+by≤1.

Appendix: Ptolemy's theorem: circle in the circle is connected to the product of the two diagonal lines of quadrilateral is equal to the sum of the products of the two pairs of opposite sides.

As shown in the figure below, ABCD is a circle connected to a quadrilateral, then the product of diagonal AC and BD is equal to the product of one pair of opposite sides AB and CD plus the product of another pair of opposite sides AD and BC, that is,

AC·BD=AB·CD+AD·BC.