Question

a,b,c属于正实数，a^2+b^2+c^2+abc=4,求证:a+b+c≤3

Answer 1

(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)
4-abc+2ab+2ac+2bc=(a+b+c)^2
要a+b+c<=3
只需√(4-abc+2ab+2ac+2bc)<=3
只需2ab+2ac+2bc-abc<=5
只需2ab+2ac+2bc-abc<=2a^2+2b^2+2c^2+2abc-3
只需(a-b)^2+(a-c)^2+(b-c)^2=>-3(abc+1)
左边恒>=0
abc>0
-3(abc+1)<0
所以就证明了a+b+c<=3