Question

la exercițiul 6 am nevoie de ajutor, dau coronița​

Answer 1

a)

a²+b²>=2ab

a²-2ab+b²>=0

(a-b)²>=0 (A)

a²+b²>= -2ab

a²+2ab+b²>=0

(a+b)²>=0 (A)

b)

a²+b²+c²>=ab+bc+ca /×2

2a²+2b²+2c²>=2ab+2bc+2ca

a²-2ab+b²+b²-2bc+c²+c²-2ca+c²>=0

(a-b)²+(b-c)²+(c-a)²>=0 (A)

c)

a+b+c>=√ab+√bc+√ca /²

a²+b²+c²>=ab+bc+ca /×2

2a²+2b²+2c²>=2ab+2bc+2ca

a²-2ab+b²+b²-2bc+c²+c²-2ca+c²>=0

(a-b)²+(b-c)²+(c-a)²=0 (A)

d)

(a+b+c)²<=3(a²+b²+c²)

a²+b²+c²+2ab+2bc+2ac<=3a²+3b²+3c²

a²+b²+c²+2ab+2bc+2ac-3a²-3b²-3c²<=0

-2a²-2b²-2c²+2ab+2bc+2ac<=0

-a²+2ab-b²-b²+2bc-c²-c²+2ca-a²<=0

-(a²-2ab+b²)-(b²-2bc+c²)-(c²-2ca+a²)

-(a-b)²-(b-c)²-(c-a)²<=0 (A)