Given that a + B + C = 1, a, B, C are positive numbers, it is proved that a ^ 2 + B ^ 2 + C ≥ 1 / 3

Given that a + B + C = 1, a, B, C are positive numbers, it is proved that a ^ 2 + B ^ 2 + C ≥ 1 / 3

How to prove a ^ 3 + B ^ 3 + C ^ 3 > = 1 / 3 (a ^ 2 + B ^ 2 + C ^ 2) (a + B + C)

Method 1
Let a > = b > = C. then use Chebyshev inequality
Method 2
If you want to prove the original form, you need to prove it
3(a^3+b^3+c^3)>=(a^2+b^2+c^2)(a+b+c)
That is, 3 (a ^ 3 + B ^ 3 + C ^ 3) > = a ^ 3 + B ^ 3 + C ^ 3 + (B + C) a ^ 2 + (a + C) B ^ 2 + (a + b) C ^ 2
That is 2 (a ^ 3 + B ^ 3 + C ^ 3) > = (B + C) a ^ 2 + (a + C) B ^ 2 + (a + b) C ^ 2.. (*)
Now let's construct an inequality
Obviously
(a + b) (a-b) ^ 2 > = 0
Namely
a^3-ab^2-ba^2+b^3>=0
Namely
a^3+b^3>=ab^2+ba^2 .(1)
In the same way
(a+c)(a-c)^2>=0
(c+b)(c-b)^2>=0
Namely
a^3+c^3>=ac^2+ca^2 .(2)
c^3+b^3>=cb^2+bc^2 .(3)
(1) + (2) + (3) learn that (*) holds
That is, the original inequality holds

It is known that a (- 1. - 1) B (1.3) C (2.5) proves that a B C is collinear

The slope of line AB is
(3+1)/(1+1)=2
The straight line is
y+1=2(x+1)
Put C in, you'll have to
Left = 5 + 1 = 6
Right = 2 × (2 + 1) = 6
equal
therefore
Collinear

Given a = K + 3, B = 2K + 2, C = 3K + 1. Find the value of a & # 178; + B & # 178; + C & # 178; + 2ab-2bc-2ac

The original formula = (a + b) &# 178; - 2C (a + b) + C & # 178;
=(a+b-c)²
=(k+3+2k+2-3k-1)²
=25

Given a = K + 3, B = 2K + 2, C = 3K-1. Find the value of a + B + C + 2ab-2bc-2ac

A + B + C + 2ab-2bc-2ac = (a + B-C) ^ 2 = (K + 3 + 2K + 2-3K + 1) ^ 2 = 6 ^ 2 = 36 hope RZ adopt!

Known: a = 1 / 3K-1, B = 1 / 6k-2, C = 1 / 2K + 3, find the square of a + the square of B + the square of C + 2ab-2ac-2bc

The square of a + the square of B + the square of C + the value of 2ab-2ac-2bc = (a + B-C) ^ 2 = (1 / 3K-1 + 1 / 6k-2-1 / 2k-3) ^ 2 = 6 ^ 2 = 36

If (2k2-3k-2) + (k2-2k) I is a pure imaginary number, then the value of real number k is equal to______ ．

If (2k2-3k-2) + (k2-2k) I is a pure imaginary number, then 2k2-3k-2 = 0 and k2-2k ≠ 0, the solution is k = - 12, so the answer is: - 12

Can you decompose a & sup2; + B & sup2; + C & sup2; + 2Ab + 2BC + 2Ac?

a²+b²+c²+2ab+2bc+2ac=(a+b+c)²

Let P = {3, log2a}, q = {a, B}, if P ∩ q = {0}, then p ∪ q = ()
A. {3，0}B. {3，0，1}C. {3，0，2}D. {3，0，1，2}

∩ P ∩ q = {0}, ∪ log2a = 0 ∪ a = 1, so B = 0, P ∪ q = {3, 0, 1}, so B

Let u = {0,1,2,3,4,5} a = {0,3,4} B = {1,3,4,5} find ① a ∩ B ② a ∪ B ③ C
Let u = {0,1,2,3,4,5} a = {0,3,4} B = {1,3,4,5}
Find ① a ∩ B ② a ∪ B ③ CUA ④ cub

A∩B=｛3,4｝
A∪B=｛0,1,3,4,5｝
CuA=｛1,2,5｝
CuB=｛0,2｝