Inequality 7320

Let a, b, and c be positive real numbers whose sum is 3, each of which is at most 2.

Prove that the inequality holds:

a2 + b2 + c2 + 3abc <9

Correct answer:

d =  1

Step-by-step explanation:

a+b+c=3 0<a<2 0<b<2 0<c<2 (a+b+c)2=a2+b2+c2+2ab+2bc+2ca (a+b+c)2=32=9  a2+b2+c2=92(ab+bc+ac) 92(ab+bc+ac)+3abc<9 2(ab+bc+ac)<3abc  2(ab+bc+ac)>3abc a=b=c=2 x11=2 (2 2+2 2+2 2)=24 x12=3 2 2 2=24 x11=x12  a=b=c=3/2 x21=2 (1.5 1.5+1.5 1.5+1.5 1.5)=13.5 x22=3 1.5 1.5 1.5=10.125 x21>x22  d=1



Did you find an error or inaccuracy? Feel free to write us. Thank you!







You need to know the following knowledge to solve this word math problem:

Related math problems and questions: