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Ta co: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left[\left(a^2+b^2+c^2\right)-ab-ac-bc\right]\\ \)
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\(B=\dfrac{\left(a+b+c\right)\left[\left(a^2+b^2+c^2\right)-ab-ac-bc\right]}{\left[\left(a^2+b^2+c^2\right)-ab-ac-bc\right]}=\left(a+b+c\right)=2016\)
\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt Cô Si: \(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
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\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt CÔ si
\(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
.............
ta có:\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
=>\(\left[\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right].\left(a+b+c\right)=a+b+c\)
=>\(\dfrac{a^2}{c+b}+\dfrac{ab}{a+c}+\dfrac{ac}{a+b}+\dfrac{b^2}{a+c}+\dfrac{ba}{c+d}+\dfrac{bc}{a+b}+\dfrac{ca}{c+d}+\dfrac{cb}{a+c}+\dfrac{c^2}{a+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{c+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
chúc bạn học tốt ^ ^
\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
+ TH1 : a + b + c = 0 ta có :
\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}\)
\(=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
+ TH2 : \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Khi đó : \(A=\left(1+1\right)\cdot\left(1+1\right)\cdot\left(1+1\right)=8\)
Ta có:
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow\left(a^3+b^3\right)+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3+3abc=0\)
\(\Rightarrow[\left(a+b\right)^3+c^3]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)[\left(a+b\right)^2-\left(a+b\right)c+c^2]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ac=0\left(2\right)\end{cases}}\)
Từ (1) => a = b = c (vì a ; b ; c là các số dương)
Giải (2) ta có:
\(2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow2a^2+2b^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Vì \(\left(a-b\right)^2\ge\forall a,b\)
\(\left(a-c\right)^2\ge\forall a,c\)
\(\left(b-c\right)^2\ge\forall b,c\)
\(\Rightarrow\)Ta có: \(a-b=a-c=b-c\Rightarrow a=b=c\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{1+1+1}=\dfrac{\left(\dfrac{3}{2}\right)^2}{3}=\dfrac{9}{\dfrac{4}{3}}=\dfrac{9}{12}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{2}\)