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

hey you, còn câu b,c?

ở đây có ai thích sơn tùng không ?

Giải:

Ta có:

\(\left(a+b+c+d\right)^2=\) \(\left[\left(a+c\right)+\left(b+d\right)\right]^2\)

\(\ge4\left(a+c\right)\left(b+d\right)\) \(=4\left(ab+bc+cd+da\right)\)\(=4\)

\(\Leftrightarrow a+b+c+d\) \(\ge2\left(a,b,c,d>0\right)\)

\(\Rightarrow\dfrac{a^3}{b+c+d}+\dfrac{b+c+d}{8}\) \(+\dfrac{b}{6}+\dfrac{1}{12}\ge\dfrac{2a}{3}\)

Tương tự ta cũng có:

\(\dfrac{b^3}{a+c+d}+\dfrac{a+c+d}{8}+\dfrac{b}{6}+\dfrac{1}{12}\) \(\ge\dfrac{2b}{3}\)

\(\dfrac{c^3}{a+b+d}+\dfrac{a+b+d}{8}+\dfrac{c}{6}+\dfrac{1}{12}\) \(\ge\dfrac{2c}{3}\)

\(\dfrac{d^3}{a+b+c}+\dfrac{a+b+c}{8}+\dfrac{d}{6}+\dfrac{1}{12}\) \(\ge\dfrac{2d}{3}\)

Cộng vế theo vế các BĐT trên ta có:

\(P\ge\dfrac{a+b+c+d}{3}-\dfrac{1}{3}\ge\) \(\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d=\dfrac{1}{2}\)

Câu hỏi của ✰✰ βєsէ ℱƐƝƝIƘ ✰✰ - Toán lớp 8 - Học toán với OnlineMath

\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)

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

\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)

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

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

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

Giải:

Từ \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\)

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

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

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

\(=-3bd\left(b+d\right)+3ac\left(b+d\right)=3\left(ac-bd\right)\left(b+d\right)=VP\) (Đpcm)

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a+b+c+d=0↔a+b=−(c+d)a+b+c+d=0↔a+b=−(c+d)

↔(a+b)³=−(c+d)³↔(a+b)³+(c+d)³=0 ↔ (a+b)³=−(c+d)³↔(a+b)³+(c+d)³=0

↔a³+b³+c³+d³+3(a+b)ab+3(c+d)cd=0 ↔ a³+b³+c³+d³+3(a+b)ab+3(c+d)cd=0

↔a³+b³+c³+d³= 3(c+d)ab−3cd(c+d)= 3(c+d)(ab−cd) ↔ a³+b³+c³+d³= 3(c+d)ab −3cd(c+d)= 3(c+d)(ab−cd)
 

Từ  \(a+b+c+d=0\)  \(\Rightarrow\) \(a+b=-\left(c+d\right)\) \(\Rightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

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

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

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

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