Cho các số thực a,b,c tm 0a,b,c1.
Cmr 2(a3+b3+c3)3+a2b+b2c+c2a


Từ gt:⇒(1a2)(1b)01+a2ba2+b
Thiết lập như vậy với 2 cái nữa rồi cộng lại:
3+a2b+b2c+c2aa2+b2+c2+a+b+c(1)
Tiếp tục:a(1+2a)(1a)0a2+a2a3
Thiết lập tương tự:a2+b2+c2+a+b+c2(a3+b3+c3)(1)
từ 1;2 suy ra đpcm
Do:a,b,c[0;1]a3a2a;b3b2b;c3c2c
suy ra:2(a3+b3+c3)a2+a+b2+b+c2+c
2(a3+b3+c3)a2bb2cc2aa2+b2+c2+a+b+c(a2b+b2c+c2a)
Cần CM:a2+b2+c2+a+b+c(a2b+b2c+c2a)3
(a21)(1b)+(b21)(1c)+(c21)(1a)0(bđt này luôn đúng)
đpcm

giả sử c=min{a,b,c}
ta đc:2c3b2c+c2a2c3b2cc2a0
vậy ta cần chứng minh:2(a3+b3)a2b3
nếu:ab0 ta có:2(a3+b3)a2b=2a3+b3+b(b2a2)2+1+0=3
nếu:ba0 ta có:2(a3+b3)=a3+2b3+a2(ab)1+2+0=3
xong,dấu = xảy ra khi a=b=c=1hoca=b=1;c=0,hoán vị các số cho nhau
a2b(1b)+b2c(1c)+c2a(1a)0
Nên a2b+b2c+c2a+3a2b2+b2c2+c2a2+3
a2b2+b2c2+c2a2+3=(1a2)(1b2)+(1b2)(1c2)+(1c2)(1a2)+2(a2+b2+c2)2(a2+b2+c2)

Suy ra a2b+b2c+c2a+32(a2+b2+c2)=2[a3+b3+c3+a2(1a)+b2(1b)+c2(1b)]2(a3+b3+c3)

Cách bạn dài quá thoạt đầu nhìn hơi khó hiểu –  tranquynhat2002 30-06-17 06:14 AM
0a,b,c1
a3+b3+c31+1+1=3
Ta cần cm 
2(a3+b3+c3)3+a2b+b2c+c2a
a2(ab)+b2(bc)+c2(ca)0
Thật vậy 
a2(ab)+b2(bc)+c2(ca)ab+bc+ca=0đpcm
Dấu bằng xảy raa=b=c=1
2(ab)+b2(bc)+ca2(ab)+b2(bc)+c2(ca0

Bạn cần đăng nhập để có thể gửi đáp án

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