Đề : ab + 4bc + ca \(\le\)0 

Có : a + b + c = 0 => a = - b - c

Thay vào ab + 4bc + ca \(\le\)0 ta đc:

(-b - c).b + 4bc + c.(-b - c) \(\le\) 0

=> -b² - bc + 4bc - bc - c² \(\le\)0

=> -b² - c² + 2bc \(\le\)0

=> - (b² - 2bc + c²) \(\le\) 0

=> -(b - c)² \(\le\) 0 (luôn đúng)

Vậy ab + 4bc + ca  \(\le\) 0

abc bằng 0

Ta có: a + b + c = 0.

=> a = - b - c

b = -a - c

c = - a- b.

Nên ta có:

ab + bc + ca = (-b-c)b + (-a-c)c + (-a-b)a

= -b^2 - bc - ca  -c^2 - a^2 - ab

= -( a^2 + b^2 + c^2)- (ab + bc + ca)

=> 2(ab + bc + ca) = -(a^2 + b^2 +c^2)

Mà -(a^2 + b^2 + c^2) bé hơn hoặc bằng 0 (do a^2 + b^2 + c^2 lớn hơn hoặc bằng 0)

=> 2(ab + bc + ca ) bé hơn hoặc bằng 0.

=> ab + bc + ca bé hơn hoặc bằng 0.

Vậy ab + bc + ca bé hơn hoặc bằng 0.

Ta có:

\(\Rightarrow a\left(a+b+c\right)=b\left(a+b+c\right)=c\left(a+b+c\right)=0\)

\(\Rightarrow a^2+ab+ac=ab+b^2+bc=ca+cb+c^2=0\)

\(\Rightarrow\left(ab+bc+ca\right)+\left(a^2+b^2+c^2\right)=0\)

Do \(a^2+b^2+c^2\ge0\Rightarrow ab+bc+ca\le0^{đpcm}\)

Áp dụng bđt Cauchy: \(3\cdot\left(ab+bc+ca\right)\le\left(a+b+c\right)^2=0\)

\(\Rightarrow ab+bc+ca\le0\)

Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

\(\Rightarrow ab+bc+ac=-\frac{1}{2}\left(a^2+b^2+c^2\right)\)

Vì \(a^2+b^2+c^2\ge0\)  \(\forall a;b;c\)

\(\Rightarrow-\frac{1}{2}\left(a^2+b^2+c^2\right)\le0\)  \(\forall a;b;c\)

Hay \(ab+bc+ac\le0\) (đpcm)

ab + bc + ca<= 0  thì a10 +b10 + c10+(b+c+a)

ab + bc + ac \(\le\)0 (đpcm)

  Ủng hộ tớ đi!

Ta có \(a+b+c=0\)

\(=>a=-b-c\)

Ta có \(ab+bc+ac\le0\)

\(=>\left(-b-c\right)b+bc+\left(-b-c\right)c\le0\)

\(=>-b^2-bc+bc-bc-c^2\le0\)

\(=>-b^2-bc-c^2\le0\)

\(=>-\left(b^2+bc+c^2\right)\le0\)(ĐPCM)

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(a^2+b^2+c^2\ge0\)

\(a^2+b^2+c^2=-\left(2ab+2bc+2ac\right)\)

\(\Rightarrow2ab+2bc+2ca\le0\Leftrightarrow ab+bc+ac\le0\)