A proof of an inequality, a proof of a mathematical inequality

On the left, only 3(A 2+B 2+C 2)-(A+B+C) 2>=03A 2+3B 2+3C 2-(A 2+B 2+C 2+2AB+2AC+2BC)=A 2+B 2+2AB+A 2+C 2+2AC+B 2+C 2+2BC=(A+B) 2+(A+C) 2+(B+C) 2>=0

When a=b=c=0, an equal sign is taken.

On the right, only the certificate (A+B+C) 2-3(AB+BC+CA)>=0A2+B2+C 2+2AB+2AC+2BC)-3(AB+BC+CA)=(A-B) 2+(A-C) 2+(B-C) 2>=0.

First (x-y) 2>=0, we get: x 2+y 2>=2xy. Let's use this inequality to prove the conclusion:

a^2+b^2>=2ab

b^2+c^2>=2bc

c^2+a^2>=2ca

The sum of the three formulas is obtained;

2(a 2 + b 2 + c 2) > = 2ab + 2bc + 2ca (1) add a 2 + b 2 + c 2 on both sides at the same time to obtain:

3(a^2+b^2+c^2)>=a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2

This proves the first inequality.

Similarly, add ab+bc+ca on both sides of equation (1) to get :

2(a^2+b^2+c^2)+ab+bc+ca>=3(ab+bc+ca)

i.e.: a+b+c) 2>=3(ab+bc+ca).

( 3a^4)-(4a^3b)+ b^4)= ( 3a^4)-(3a^3b)-a^3b+ （b^4)=3a^3(a-b)-b(a^3-b^3)=3a^3(a-b)-b(a-b)(a^2+ab+b^2)=(a-b)(3a^3-ba^2-ab^2-b^3)=(a-b)[(a^3-b^3)+a^2(a-b)+a(a^2-b^2)]

a-b) 2(3a 2+2ab+b 2)=(a-b) 2[2a 2+(a+b) 2](3a 4)-(4a 3b) + b 4) is greater than or equal to 0 because (a-b) 2, a 2, (a+b) 2 are all greater than or equal to 0

Anti-beam chain method: assuming x+y>2, then the old y>2-xx 3+y 3>x 3+(2-x) 3=8-12x+6x 2=6(x-1) 2+2 2

i.e. x 3 + y 3 > 2 and the inscription x 3 + y 3 = 2 spear early rise shield.

So the assumption is not true, x+y<=2

Because for any real number x,y there is an inequality 2(x 2+y 2)>=(x+y) 2 holds, so.

A 2+b 2>=(1 2)(a+b) 2,c 2+d 2>=(1 2)(c+d) 2, therefore.

a^2+b^2+c^2+d^2

(1/2)(a+b)^2+(1/2)(c+d)^2=(1/2)[(a+b)^2+(c+d)^2]>=(1/2)*(1/2)*(a+b+c+d)^2=1/4

i.e. a 2 + b 2 + c 2 + d 2>=1 4

Because 3x 2+y 2+z 2-2x(y+z)=x 2+(x 2+y 2-2xy)+(x 2+z 2-2xz)=x 2+(x-y) 2+(x-z) 2=0, x=0, x-y=0, x-z=0, i.e., x=y=z=0