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 :

\(a,b,c>0\) \(\Rightarrow\) \(ab+bc+ca>0\)mâu thuẫn với đề bài 

Xem lại đề..

Ta có: a,b,c > 0 => a+b+c=0 thì ab+bc+ca\(\le0\)

a+b+c=0 => (a+b+c)²=0

=> a²+b²+c²+2(ab+bc+ca)=0

=> 2(ab+bc+ca)=-(a²+b²+c²)

Mà \(a^2+b^2+c^2\ge0\Rightarrow-\left(a^2+b^2+c^2\right)\le0\)

Do đó \(2\left(ab+bc+ca\right)\le0\)

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

\(\frac{\left(2-c\right)\left(b-c\right)}{2a+bc}=\frac{\left(a+b\right)\left(b-c\right)}{a\left(a+b+c\right)+bc}=\frac{\left(a+b\right)\left(b-c\right)}{\left(a+b\right)\left(c+a\right)}=\frac{b-c}{c+a}=\frac{b}{c+a}-\frac{c}{c+a}\)

Tương tự, ta có: \(\frac{\left(2-a\right)\left(c-a\right)}{2b+ca}=\frac{c}{a+b}-\frac{a}{a+b};\frac{\left(2-b\right)\left(a-b\right)}{2c+ab}=\frac{a}{b+c}-\frac{b}{b+c}\)

\(\Rightarrow\)\(VT=\left(\frac{a}{b+c}-\frac{a}{a+b}\right)+\left(\frac{b}{c+a}-\frac{b}{b+c}\right)+\left(\frac{c}{a+b}-\frac{c}{c+a}\right)\)

\(=\frac{a\left(a-c\right)}{\left(a+b\right)\left(b+c\right)}+\frac{b\left(b-a\right)}{\left(b+c\right)\left(c+a\right)}+\frac{c\left(c-b\right)}{\left(c+a\right)\left(a+b\right)}\)

\(=\frac{a\left(a-c\right)\left(c+a\right)+b\left(b-a\right)\left(a+b\right)+c\left(c-b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(=\frac{\left(a^3+b^3+c^3\right)-\left(a^2b+b^2c+c^2a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{\left(a^3+b^3+c^3\right)-\left(a^3+b^3+c^3\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{3}\)

cái bđt \(a^3+b^3+c^3\ge a^2b+b^2c+c^2a\) cô Chi có làm r ib mk gửi link 

1)Cho a,b,c >0

Chứng minh  bc/a^2(b+c) + ca/b^2(c+a) +ab/c^2(a+b) > hoặc = 1/2(1/a+1/b+1/c)

2) Cho a,b,c>0 1/a + 1/b + 1/c =1

Chứng minh (b+c)/a^2 + (c+a)/b^2 + (a+b)/c^2 > hoặc = 2

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