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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)
Ta có: a+b+c+d=0
⇔\(a+d=-\left(b+c\right)\)
\(\Leftrightarrow\left(a+d\right)^3=-\left(b+c\right)^3\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-\left[b^3+c^3+3bc\left(b+c\right)\right]\)
\(\Leftrightarrow a^3+d^3+3ad\left(a+d\right)=-b^3-c^3-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+d^3+b^3+c^3=-3ad\left(a+d\right)-3bc\left(b+c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ad\left(a+d\right)+3bc\left(a+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\left(-3ad+3bc\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\cdot3\cdot\left(-ad+bc\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-\left(b+c\right)\cdot3\cdot\left[-\left(ad-bc\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\cdot\left(b+c\right)\cdot\left(ad-bc\right)\)(đpcm)
Ta có :
\(a+b+c+d=0\)
\(\Rightarrow b+c=-\left(a+d\right)\)
\(\Rightarrow\left(b+c\right)^2=\left(a+d\right)^2\)
\(\Rightarrow\left(b+c\right)^2-\left(a+d\right)^2=0\)
\(\Rightarrow b^2+c^2+2bc-a^2-d^2-2ad=0\)
Lại có :
\(a^3+b^3+c^3+d^3\)
\(=\left(a+d\right)\left(a^2+d^2-ad\right)+\left(b+c\right)\left(b^2+c^2-bc\right)\)
\(=\left(b+c\right)\left(b^2+c^2-bc\right)-\left(b+c\right)\left(a^2+d^2-ad\right)\)
\(=\left(b+c\right)\left[\left(b^2+c^2-bc\right)-\left(a^2+d^2-ad\right)\right]\)
\(=\left(b+c\right)\left[\left(b^2+c^2+2bc-a^2-d^2-2ad\right)+3ad-3bc\right]\)
\(=\left(b+c\right)\left[0+3\left(ad-bc\right)\right]\)
\(=3\left(b+c\right)\left(ad-bc\right)\)
Vậy ...
Ta có : a + b +c + d = 0
=> a + d = - b - c
=> (a + d) = -(b + c)
=> (a + d)3 = -(b + c)3
a3 + 3a2d + 3ad2 + d3 = -(b3 + 3b2c + 3bc2 + c3)
a3 + 3a2d + 3ad2 + d3 = -b3 - 3b2c - 3bc2 - c3
a3 + b3 + c3 + d3 = -3a2d - 3ad2 - 3b2c - 3bc2
a3 + b3 + c3 + d3 = -3ad(a + d) - 3bc(b + c)
a3 + b3 + c3 + d3 = -3ad(-b - c) - 3bc(b + c)
a3 + b3 + c3 + d3 = 3ad(b + c) - 3bc(b + c)
a3 + b3 + c3 + d3 = 3(b + c)(ad - bc)
Theo đề, a+b+c+d=0
\(\Rightarrow a+b=-\left(c+d\right)\)
Ta có: \(VT=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(c+d\right)\left(c^2-cd+d^2\right)\)
\(\Leftrightarrow VT=\left(c+d)\left(c^2-cd+d^2-a^2+ab-b^2\right)\right)\)
Để có ĐPCM ta xét hiệu: \(c^2-cd+d^2-a^2+ab-b^2-3\left(ab+cd\right)=c^2-4cd+d^2-a^2-2ab-b^2=c^2-4cd+d^2-\left(a+b\right)^2=c^2-4cd+d^2-\left(c+d\right)^2=-6cd\)
S nó ko = 0 ta:::xem lại đề..Hay mk lm sai j đó
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c