Lời giải:
$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$

$=[(a+b+c)-2(ab+bc+ac)]^2-2(a^2b^2+b^2c^2+c^2a^2)$

$=[-2(ab+bc+ac)]^2-2(a^2b^2+b^2c^2+c^2a^2)$

$=4(ab+bc+ac)^2-2[(ab+bc+ac)^2-2abc(a+b+c)]$

$=4(ab+bc+ac)^2-2[(ab+bc+ac)^2]=2(ab+bc+ac)^2$
Ta có đpcm.

Ta có :

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

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

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

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

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

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\left(1\right)\)

\(\Leftrightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\) (vì \(a+b+c=0\))

\(\left(1\right)+\left(2\right)\Rightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)

\(\Rightarrow dpcm\)

a + b + c = 0

=> (a + b + c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

=> a² + b² + c² = -2(ab + 2bc + 2ca)

=> (a² + b² + c²)² = [-2(ab + bc + ca)]²

=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 4(a²b² + b²c² + c²a² + 2ab²c + 2a²bc + 2abc²) 

=> a⁴ + b⁴ + c⁴ = 4a²b² + 4b²c² + 4c²a² + 8a²bc + 8ab²c + 8abc² - 2a²b² - 2b²c² - 2a²c²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 8abc(a + b + c)

=> a⁴ + b⁴ + c⁴= 2a²b² + 2b²c² + c²a²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 2abc(a + b + c) (Vì a + b + c = 0)

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 2a²bc + 2ab²c + 2abc²

=>  a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + c²a² + a²bc + ab²c + abc²) 

=> a⁴ + b⁴ + c⁴ = 2(ab + bc + ca)² (đpcm)

Ta có​ : \(a+b+c=0\)

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

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

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

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

Bình phương hai vế , ta được :

\(\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ac\right)\right]^2\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\right)\) \(\left(1\right)\)

\(\Rightarrow a^4+b^4+c^4=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\) ( vì \(a+b+c=0\) ) \(\left(2\right)\)

Từ​ \(\left(1\right),\left(2\right)\):

\(2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)

Ta có : a + b + c = 0

( a + b + c )\(^2\) = 0

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

Nên : \(a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(ab+bc+ca\right)^2\)

\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\right)\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+8ab^2c+8abc^2+8a^2bc\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+8abc\left(b+c+a\right)\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)

Lại có : \(2\left(ab+bc+ca\right)^2\)

\(=2\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\right)\)

\(=2a^2b^2+2b^2c^2+2c^2a^2+4ab^2c+4abc^2+4a^2bc\)

\(=2a^2b^2+2b^2c^2+2c^2a^2+4abc\left(b+c+a\right)\)

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

Vì : \(2a^2b^2+2b^2c^2+2c^2a^2=2a^2b^2+2b^2c^2=2c^2a^2\)

Vậy \(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)

a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)

Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)

\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)

b. 

\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)

\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)

\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)

- TH1: nếu \(a+b+c\ge4\)

Ta có: \(ab+bc+ca=4-abc\le4\)

\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)

(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)

- TH2: nếu \(3\le a+b+c< 4\)

Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)

\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)

Áp dụng BĐT Schur:

\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)

\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)

(Dấu "=" xảy ra khi \(a=b=c=1\))

\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+abc^2+a^2bc\right)=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow2\left(ab^2c+abc^2+a^2bc\right)=0\\ \Leftrightarrow abc\left(a+b+c\right)=0\left(đpcm;a+b+c=0\right)\)

Ta có: \(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2=c^2\)

\(\Rightarrow a^2+b^2-c^2=-2ab\)

\(\Rightarrow\left(a^2+b^2-c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2-b^2c^2-c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=\left(-2ab\right)^2-2a^2b^2+2b^2c^2+2c^2a^2=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(đpcm\right)\)

Dùng hằng đang thuc la ra~~~daif qua nen ngai viet

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