cmr nếu a+b+c=(căn)ab + (căn)bc + (căn)ac thì a=b=c 16/08/2021 Bởi Kinsley cmr nếu a+b+c=(căn)ab + (căn)bc + (căn)ac thì a=b=c
a+b >= 2 căn(ab) b+c >= 2 căn(bc) c+a >= 2 căn(ac) => a+b+b+c+c+a >= 2 căn(ab) + 2 căn(bc) + 2 căn(ac) => 2(a+b+c) >= 2((căn)ab + (căn)bc + (căn)ac)=> a+b+c >= (căn)ab + (căn)bc + (căn)ac dấu “=” xảy ra khi a=b=c Bình luận
a+b+c= $\sqrt{ab}$ + $\sqrt{bc}$+ $\sqrt{ac}$ => 2a+ 2b+ 2c= $\sqrt{ab}$ + $\sqrt{bc}$+ $\sqrt{ac}$ <=> 2a+ 2b+ 2c- 2$\sqrt{ab}$ – 2$\sqrt{bc}$- 2$\sqrt{ac}$ =0 <=> (√a- 2$\sqrt{ab}$+ √b)+ (√b- 2$\sqrt{bc}$+ √c)+ (√a- 2$\sqrt{ac}$+ √c)=0 <=> (√a- √b)²+ (√b- √c)²+ (√a- √c)²=0 Dấu “=” xảy ra <=> √a= √b= √c <=> a=b=c Bình luận
a+b >= 2 căn(ab)
b+c >= 2 căn(bc)
c+a >= 2 căn(ac)
=> a+b+b+c+c+a >= 2 căn(ab) + 2 căn(bc) + 2 căn(ac)
=> 2(a+b+c) >= 2((căn)ab + (căn)bc + (căn)ac)
=> a+b+c >= (căn)ab + (căn)bc + (căn)ac
dấu “=” xảy ra khi a=b=c
a+b+c= $\sqrt{ab}$ + $\sqrt{bc}$+ $\sqrt{ac}$
=> 2a+ 2b+ 2c= $\sqrt{ab}$ + $\sqrt{bc}$+ $\sqrt{ac}$
<=> 2a+ 2b+ 2c- 2$\sqrt{ab}$ – 2$\sqrt{bc}$- 2$\sqrt{ac}$ =0
<=> (√a- 2$\sqrt{ab}$+ √b)+ (√b- 2$\sqrt{bc}$+ √c)+ (√a- 2$\sqrt{ac}$+ √c)=0
<=> (√a- √b)²+ (√b- √c)²+ (√a- √c)²=0
Dấu “=” xảy ra <=> √a= √b= √c <=> a=b=c