\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow ab+bc+ca=0\)

\(a+b+c=\sqrt{2019}\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2019\)

\(\Rightarrow a^2+b^2+c^2=2019\) ( vì \(ab+bc+ca=0\))

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\\ A=a^2+b^2+c^2\\ \Leftrightarrow A=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\\ \Leftrightarrow A=\left(\sqrt{2019}\right)^2-2\cdot0=2019\)

Theo bất đẳng thức tam giác

\(\Rightarrow\left\{\begin{matrix}a< b+c\\b< c+a\\c< a+b\end{matrix}\right.\Rightarrow\left\{\begin{matrix}b+c-a>0\\c+a-b>0\\a+b-c>0\end{matrix}\right.\)

Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a,b>0\)

\(\Rightarrow\left\{\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{a+c-b}\ge\dfrac{2}{a}\end{matrix}\right.\)

Cộng theo từng vế

\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( đpcm )

câu 1: a+b>?

Câu 1: mik sửa đề tí

Ta có: a+b=1

a² +b² ≥ (a+b)²/2

<=> a² +b² ≥ 1/2(a² +b²) + ab

<=> 1/2(a² +b²) -ab ≥ 0

<=> 1/2(a-b)² ≥ 0 ( luôn đúng )

vậy a² + b² ≥ (a+b)²/2 = 1/2

tương tự thì

a^4 + b^4 ≥ (a² +b²)²/2 ≥ (1/2)²/2 = 1/8

vậy a^4 + b^4 ≥ 1/8

dấu = xảy ra <=> a=b=1/2

\(B=\Sigma\frac{ab}{a^2+b^2-c^2}\)

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

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

\(B=\frac{ab}{a\left(a-b+c\right)}+\frac{bc}{b\left(b-c+a\right)}+\frac{ac}{c\left(c-a+b\right)}\)

\(B=\frac{b}{a+b+c-2b}+\frac{c}{a+b+c-2c}+\frac{a}{a+b+c-2a}\)

\(B=\frac{-b}{2b}+\frac{-c}{2c}+\frac{-a}{2a}\)

\(B=\frac{-1}{2}+\frac{-1}{2}+\frac{-1}{2}\)

\(B=\frac{-3}{2}\)

+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Rightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)

\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)

Ta có: \(a^2+b^2+c^2=1\)

⇒ \(\left\{{}\begin{matrix}\left|a\right|\text{≤}1\\\left|b\right|\text{≤}1\\\left|c\right|\text{≤}1\end{matrix}\right.\)

Mặt khác:

\(a^2+b^2+c^2=a^3+b^3+c^3=1\)

⇒ \(a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)=0\)

Vì \(\left\{{}\begin{matrix}1-a\text{≥}0\\1-b\text{≥}0\\1-c\text{≥}0\end{matrix}\right.\) 

⇒ \(a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\text{≥}0\)

Dấu "=" ⇔ 1 số bằng 1 và 2 số còn lại bằng 0

⇒ \(S=1\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2016}\)

\(\Rightarrow\dfrac{bc+ac+bc}{abc}=\dfrac{1}{2016}\)

\(\Rightarrow\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc=abc\)

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

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

\(\Rightarrow a=-b\) hay \(b=-c\) hay \(c=-a\)
-Vậy trong ba số a,b,c tồn tại 2 số đối nhau.

Ta có a+b+c=0

<=> a+b=-c <=>a²+b²-c²=-2ab

   b+c=-a <=> b²+c²-a²=-2bc

  c+a=-b <=> c²+a²-b²=-2ca

Thay vào biểu thức ta có

\(B=\frac{ab}{-2ab}-\frac{bc}{2bc}-\frac{ca}{2ca}=\frac{-3}{2}\)