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.

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 :

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow dpcm\)

Với mọi số thực dương a;b;c ta có BĐT:

\(a^4+b^4\ge ab\left(a^2+b^2\right)\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Tương tự, ta có:

\(VT\le\dfrac{ab}{ab\left(a^2+b^2\right)+ab}+\dfrac{bc}{bc\left(b^2+c^2\right)+bc}+\dfrac{ca}{ca\left(c^2+a^2\right)+ca}\)

\(VT\le\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(VT\le\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\)

Ta lại có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{xyz}{xy\left(x+y\right)+xyz}+\dfrac{xyz}{yz\left(y+z\right)+xyz}+\dfrac{xyz}{zx\left(z+x\right)+xyz}=1\)

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

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

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

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

\(=4\left(ab+bc+ca\right)^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+a^2bc+abc^2\right)\right]\)

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

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

\(=2\left(ab+bc+ca\right)^2\)