\(\dfrac{a^3}{b}+ab+\dfrac{b^3}{c}+bc+\dfrac{c^3}{a}+ca\ge2\sqrt{\dfrac{a^4b}{b}}+2\sqrt{\dfrac{b^4c}{c}}+2\sqrt{\dfrac{c^4a}{a}}=2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

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

áp dụng AM GM ta có a^3/b+ab>=2a^2

chứng minh tương tự => a^3/b+b^3/c+c^3/a>=2(a^2+b^2+c^2)-(ab+bc+ca)

mà ta có a^2+b^2+c^2>=(ab+bc+ca)

=>a^3/b+b^3/c+c^3/a>= ab+bc+ca

"=" xảy ra khi a=b=c

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+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) [vì a+b = - (c+d)] 
==> a^3 + b^^3 + c^3 + d^3 =3 (c+d) (ab-cd) (đpcm)

 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) ( Vì a+b = - (c+d)) 
==> a^3 + b^3 + c^3 + d^3 = 3(c+d)(ab-cd) (đpcm).

ta có a+b+c+d = 0=> b+c= -( a+d) => (b+c)^3 = - (a+d)^3

=> b^3+ c^3 + 3bc( b+c) = -( a^3 +d^3 + 3ad(a+d))

=> a^3+b^3+c^3+d^3 = - 3ad( a+d) - 3bc(b+c) = 3ad(b+c) - 3bc(b+c) 

= 3(b+c)(ad-bc)

sao cậu tự đặt câu hỏi rồi lại tự trả lời luôn         

       thế là sao??????????

minh thich la minh lam thoi

a+b+c+d=0

=>a+d=-(b+c)

=>(a+d)^3=-(b+c)^3

=>\(a^3+d^3+3ad\left(a+d\right)=-b^3-c^3-3bc\left(b+c\right)\)

=>\(a^3+d^3+3ad\left(a+d\right)=-b^3-c^3+3bc\left(a+d\right)\)

=>\(a^3+d^3+b^3+c^3=3bc\left(a+d\right)-3ad\left(a+d\right)\)

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

=>\(a^3+b^3+c^3+d^3=3\left(b+c\right)\left(ad-bc\right)\)

Ta có: a+b+c+d=0

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

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

\(\Leftrightarrow b^3+c^3+3bc\left(b+c\right)=-\left[a^3+d^3+3ad\left(a+d\right)\right]\)

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

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bc\left(b+c\right)-3ad\cdot\left[-\left(b+c\right)\right]\)

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

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