Inequality (also called sorting principle) is the outline of high school mathematics competition and the new curriculum standard.
Ordinary high school curriculum standard test textbook (People's Education Press) Mathematics (elective 4-5
Lecture 3 (Part 3)
Basic inequalities are needed.
There are two sets of numbers.
a 1
,
Aortic second sound
,……
Ann;
b 1
,
b2
,……
billion
satisfy
a 1
≤
Aortic second sound
≤……≤
Ann,
b 1
≤
b2
≤……≤
billion
, where C 1, C2, ..., CN is B 1, B2, ..., bn, and then there is.
a 1*
billion
+
Aortic second sound
*b{n- 1}+
...
+
One; one
*b 1
≤
a 1
*c 1
+
a2*
c2}
+……+
One; one
*cn}
≤
a 1
*b 1
+
Aortic second sound
*b2
+
.....+Ann *
10 billion
The only possibility is that
a 1
=
Aortic second sound
=
...
=
One; one
or
b 1
=
b2
=
...
=
billion
The time equal sign holds, that is, the sum in reverse order is equal to the sum in positive order.
Sorting inequality is often used to describe the relationship between the products of a set of numbers that have nothing to do with order. Can shilling a 1
≤
Aortic second sound
≤
a3
≤
...
≤
Ann, determine the relationship between size.
When using, a group of numbers are often constructed and solved by forming a product relationship with the original number. It is suitable for proving fractions, products, especially rotational inequalities.
The above ranking inequality can also be abbreviated as:
Inverse order and ≤ disordered order and ≤ homologous order sum
.
Proof of rank inequality;
Step by step adjustment method.
When n=2, we might as well set a 1.
≤
a2，
b 1
≤
Then go to B2.
a 1*b 1
+
a*2
b*2
-
(
a2*b 1
+
a 1*b2)
=
(
a 1
-
Aortic second sound
)(
b 1
-
b2
)
≥0.
Therefore, n=2 holds.
When n >; 2. Only two inequalities need to be proved separately.
Let's set a 1.
≤
Aortic second sound
≤
...
≤
an，b 1
≤
b2
≤
...
≤
bn .
A.
Disordered sum and ≤ congruent sum
check
a 1
b{t_ 1}
+
Aortic second sound
b{t_2}
+
...
+
One; one
b{t_n} .
If t 1= 1, check t2. If ti=i, i= 1,
...,
K, and then examine t{k+ 1}.
Now let's set the first footnote that satisfies tk>k as m, that is, a 1.
b 1
+
Aortic second sound
b2
+
...
+
be
b{tm}
+
...
+
One; one
b{tn}，tm & gtm .
And find out the item containing b_m, and let it be a _ L.
b_m，l & gtm .
So, because in the morning
≤
a_l，b_{t_m}
≥
B_m, so a_m
b_m
+
American Airlines
b _ t _ m }
≥
morning
b _ t _ m }
+
American Airlines
b_m。
Therefore, the two items are arranged in the same order and then become larger.
The adjusted formula becomes
a_ 1
b_ 1
+
a2
B2
+
...
+
morning
b _ t _ m }
+
...
+
American Airlines
b_{t_n}
≤a_ 1
b_ 1
+
a2
B2
+
...
+
morning
b_m
+
...
+
American Airlines
b_{t_n}
Because such a term is limited, the same order sum is obtained after finite step adjustment, thus proving that the disordered sum is less than or equal to the same order sum.
B.
Inverse sum ≤ disordered sum
Completely similar to the proof of A, every step reduced can be proved by a limited number of steps.
The necessary and sufficient conditions for obtaining the equal sign are: A 1 = A2 =...= Ann.
or
b 1=b2=……bn