co a+b+c=0 =>b+c=-a

suy ra (b+c)²=(-a)²  hay b²+2bc+c² =a²

hay b²+c²-a² =-2bc

Suy ra (b² + c² - a² )² =( -2bc)²

<=> b⁴ +c⁴ +a⁴ +2b²c² -2a²b² -2a²c² = 4b²c²

<=> a⁴+b⁴+c⁴ =2a²b²+2b²c²+2c²a²

<=> 2(a⁴+b⁴+c⁴) = a⁴+b⁴+c⁴+2a²b²+2b²c²+2c²a²

<=> a²+b²+c² =2(a⁴+b⁴+c⁴) (dpcm)

Giá trị này trong nào chả bằng 0

Từ đề bài ta có :

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

Mà \(a^2+b^2+c^2=1\)  < = > 1 + 2 ( ab + ac + bc ) = 0

< = > 2 ( ab + ac + bc ) = -1 

< = > ab + ac + bc = -1/2

\(< =>\left(ab+ac+bc\right)^2=\left(-\dfrac{1}{2}\right)^2< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\)

\(< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2=\dfrac{1}{4}\)

Lại có từ \(a^2+b^2+c^2=1\)

\(< =>\left(a^2+b^2+c^2\right)^2=1< =>a^4+b^4+c^4+2\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]=1\)

\(< =>a^4+b^4+c^4+2.\dfrac{1}{4}=1< =>a^4+b^4+c^4+\dfrac{1}{2}=1< =>a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\left(đpcm\right)\)

a + b + c = 0
<=> (a + b + c)² = 0
<=> a² + b² + c² + 2(ab + bc + ca) = 0
<=> a² + b² + c² = -2(ab + bc + ca) (1)

CẦn chứng minh:

2(a^4 + b^4 + c^4) = (a² + b² + c²)²

<=> 2(a^4 + b^4 + c^4) = a^4 + b^4 + c^4 + 2(a²b² + b²c² + c²a²)

<=> a^4 + b^4 + c^4 = 2(a²b² + b²c² + c²a²)

<=> (a² + b² + c²)² = 4(a²b² + b²c² + c²a²) ---(cộng 2 vế cho 2(a²b² + b²c² + c²a²) )

<=> [-2(ab + bc + ca)]² = 4(a²b² + b²c² + c²a²) ----(do (1))

<=> 4.(a²b² + b²c² + c²a²) + 8.(ab²c + bc²a + a²bc) = 4(a²b² + b²c² + c²a²)

<=> 8.(ab²c + bc²a + a²bc) = 0

<=> 8abc.(a + b + c) = 0

<=> 0 = 0 (đúng), Vì a + b + c = 0

=> Đpcm

a + b + c = 0

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

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

\(\Leftrightarrow a^2+b^2+c^2=-2.\left(ab+bc+ca\right)\left(1\right)\)

Cần phải chứng minh

2.(a⁴ + b⁴ + c⁴)=(a²+b²+c²)

\(\Leftrightarrow\) 2.(a⁴ - b⁴+c⁴)=a⁴+b⁴+c⁴+2.(a²b²+b²c²+c²a²)

\(\Leftrightarrow\)a⁴ +b⁴+c⁴=2.(a²b²+b²c²+c²a²)

\(\Leftrightarrow\) (a² + b² +c² ) = 4(a²b²+b²c² +c²a²)

\(\Leftrightarrow\) [ -2.(ab+bc+ca)² ] = 4(a²b²+b²c² +c²a²)

\(\Leftrightarrow\) 4(a²b²+b²c² +c²a²)+8.(ab²c +bc²a+a²bc)=4.(a²b+b²c²+c²+a²

^{\(\Leftrightarrow\)} 8(ab²c+bc²a+a²bc)=0

\(\Leftrightarrow\)8abc.(a+b+c)=0

\(\Leftrightarrow\) 0 =0 (đúng ) Vì a +b +c =0

=> ĐPCM

Áp dụng  \(x^2+y^2+z^2\ge xy+yz+zx\) và \(x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^2\)

\(N\ge\dfrac{a^2b}{c}+\dfrac{b^2c}{a}+\dfrac{c^2a}{b}\ge\dfrac{1}{3}\left(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}\right)^2=3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Bài 6 . Áp dụng BĐT Cauchy , ta có :

a² + b² ≥ 2ab ( a > 0 ; b > 0)

⇔ ( a + b)² ≥ 4ab

⇔ \(\dfrac{\left(a+b\right)^2}{4}\)≥ ab

⇔ \(\dfrac{a+b}{4}\) ≥ \(\dfrac{ab}{a+b}\) ( 1 )

CMTT , ta cũng được : \(\dfrac{b+c}{4}\) ≥ \(\dfrac{bc}{b+c}\) ( 2) ; \(\dfrac{a+c}{4}\) ≥ \(\dfrac{ac}{a+c}\)( 3)

Cộng từng vế của ( 1 ; 2 ; 3 ) , Ta có :

\(\dfrac{a+b}{4}\) + \(\dfrac{b+c}{4}\) + \(\dfrac{a+c}{4}\) ≥ \(\dfrac{ab}{a+b}\) + \(\dfrac{bc}{b+c}\) + \(\dfrac{ac}{a+c}\)

⇔ \(\dfrac{a+b+c}{2}\) ≥ \(\dfrac{ab}{a+b}\) + \(\dfrac{bc}{b+c}\) + \(\dfrac{ac}{a+c}\)

Bài 4.

Áp dụng BĐT Cauchy cho các số dương a , b, c , ta có :

\(1+\dfrac{a}{b}\) ≥ \(2\sqrt{\dfrac{a}{b}}\) ( a > 0 ; b > 0) ( 1)

\(1+\dfrac{b}{c}\) ≥ \(2\sqrt{\dfrac{b}{c}}\) ( b > 0 ; c > 0) ( 2)

\(1+\dfrac{c}{a}\) ≥ \(2\sqrt{\dfrac{c}{a}}\) ( a > 0 ; c > 0) ( 3)

Nhân từng vế của ( 1 ; 2 ; 3) , ta được :

\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\) ≥ \(8\sqrt{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=8\)

@Giáo Viên Hoc24.vn

@Akai Haruma