Cách khác dễ hiểu hơn

Áp dụng BĐT Cô si 2 số ko âm 

Ta có: \(\frac{a^3}{b}+ab\ge2\sqrt{a^4}=2a^2\)

Tương tự rồi sau đó lại có:

\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+bc+ca\ge2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)

Áp dụng BĐT Cô si với 3 số k âm 

\(\frac{a^3}{b}+\frac{a^3}{b}+b^2\ge\frac{3\sqrt[3]{a^3.a^3.b^2}}{b^2}=3a^2\)

\(\frac{b^3}{c}+\frac{b^3}{c}+b^2\ge3b^2\)

\(\frac{c^3}{a}+\frac{c^3}{a}+c^2\ge3c^2\)

\(\Rightarrow2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+a^2+b^2+c^2\ge3\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)

Mà \(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)

a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b)-3bca

=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)

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

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

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

Ta có : \(a+b+c+d=0\)

\(\Leftrightarrow a+b=-c-d\)

\(\Leftrightarrow\left(a+b\right)^3=\left(-c-d\right)^3\)

\(\Leftrightarrow a^3+b^3+3ab.\left(a+b\right)=-c^3-d^3+3cd.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3cd.\left(c+d\right)-3ab.\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.cd.\left(a+b\right)+3ab.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.\left(c+d\right)\left(cd+ab\right)\)

Ta có : 

⇔a3+b3+c3+d3=3.cd.(a+b)+3ab.(c+d)

a )

`VP= (a+b)^3-3ab(a+b)`

     `=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`

     `=a^3+b^3 =VT (đpcm)`

b) 

b) Ta có

a) Ta có:

`VP= (a+b)^3-3ab(a+b)`

     `=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`

     `=a^3 + b^3=VT(dpcm)`

b) Ta có

Đặt A = a + b, B = c. Áp dụng hằng đẳng thức ( A   +   B ) 3  để biến đổi vế trái.