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\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ac\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)
Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+abc\left(a+b+c\right)\)
\(=a^2b^2+b^2c^2+a^2c^2\)
nên \(a^4+b^4+c^4=4\left(ab+bc+ac\right)^2-2\left(ab+bc+ac\right)^2\)
\(a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\left(đpcm\right)\)
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Ta có: \(a^3+b^3+c^3\ge3abc\) ( BĐT Cauchy )
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{abc}{b}+\dfrac{abc}{c}+\dfrac{abc}{a}\)
Hay \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ac+ab+bc\left(đpcm\right)\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{a^3}{b}+ab\) ≥ \(2\sqrt{\dfrac{a^3}{b}.ab}=2\sqrt{a^4}=2a^2\left(1\right)\)
\(\dfrac{b^3}{c}+bc\) ≥ \(2\sqrt{\dfrac{b^3}{c}.bc}=2\sqrt{b^4}=2b^2\left(2\right)\)
\(\dfrac{c^3}{a}+ac\) ≥ \(2\sqrt{\dfrac{c^3}{a}.ac}=2\sqrt{c^4}=2c^2\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3) , ta có :
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ac\) ≥ \(2\left(a^2+b^2+c^2\right)\) ( * )
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(a^2+b^2\) ≥ \(2ab\left(4\right)\)
\(b^2+c^2\) ≥ \(2bc\left(5\right)\)
\(c^2+a^2\) ≥ \(2ac\left(6\right)\)
Cộng từng vế của ( 4 ; 5 ; 6) , ta có :
\(2\left(a^2+b^2+c^2\right)\) ≥ \(2\left(ab+bc+ac\right)\) ( ** )
Từ ( * ; ** ) , ta có :
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ac\) ≥ \(2\left(ab+bc+ac\right)\)
⇔ \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\) ≥ \(ab+bc+ac\)
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a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (dpcm)
Cho mk nói bạn Alan Walker chỉ là hs lớp 6 sao tài vậy
Nếu bạn ko biết làm thì thôi
Làm nhục anh em bạn ạ
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\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
\(=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
\(=4\left(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ac\right)^2\)
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`(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`VT>=0`
Dấu "=" xảy ra khi `a=b=c`
`a^3+b^3+c^3=3abc`
`<=>a^3+b^3+c^3-3abc=0`
`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`**a+b+c=0`
`**a^2+b^2+c^2=ab+bc+ca`
`<=>a=b=c`
\(VT=a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\)
\(=\left(b^2c+a^2c\right)+\left(a^2b+bc^2\right)+\left(ab^2+ac^2\right)\)
bđt AM-GM:
\(VT\ge2\sqrt{a^2b^2c^2}+2\sqrt{a^2b^2c^2}+2\sqrt{a^2b^2c^2}\)
\(=2abc+2abc+2abc=6abc=VP\)
\("="\Leftrightarrow a=b=c\)