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\(a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\left(1\right)\)
\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2-a\left(b-c-d-e\right)\ge0\)
\(\Leftrightarrow\left(b^2-ab+\frac{1}{4}a^2\right)+\left(c^2-ac+\frac{1}{4}a^2\right)+\left(d^2-ad+\frac{1}{4}a^2\right)+\left(e^2-ae+\frac{1}{4}a^2\right)\ge0\)
\(\Leftrightarrow\left(b+\frac{1}{2}a\right)^2+\left(c+\frac{1}{2}a\right)^2+\left(d+\frac{1}{2}a\right)^2+\left(e+\frac{1}{2}a\right)^2\ge0\left(2\right)\)
( 2 ) đúng => ( 1 ) đúng
a, VP:-(b-a)3=-(b3-3b2a+3ba2-a3)=a3-3a2b+3ab2-b3=(a-b)3 Kết luận:VP=VT
b, VT:(-a-b)2=[(-a)+(-b)]2=(-a)2+2(-a)(-b)+(-b)2=a2+2ab+b2=(a+b)2 Kết Luận:VT=VP
\(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(-a-b\right)^2=a^2-2\left(-a\right)b+b^2\)\(=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(-a-b\right)^2\)( đpcm )
Ta có:
\(\left(-a-b\right)^2=[-\left(a+b\right)]^2=[-\left(a+b\right)]\times[-\left(a+b\right)]=\left(a+b\right)\times\left(a+b\right)=\left(a+b\right)^2\)
\(\Rightarrow\left(a+b\right)^2=\left(-a-b\right)^2\)(đpcm)
Trieu Trong Thai
CM a3+b3+c2 >= ab+bc+ac (*)
2a^2 +2b^2 +2c^2 - 2ab -2bc -2ac = (a-b)^2 + (b-c)^2 + (a-c)^2 >= 0
từ * => a^2 +b^2+c^2 +2ab+2bc+2ac >= 3ab+3bc+3ac <=> (a+b+c)^2 >= 3ab +3ac+3bc
từ * => 2ab +2ac+2bc+ a^2+b^2+c^2 =< 3a^2+3b^2+3c^2 <=> (a+b+c)^2 =< ...
de bai sai sua lai la
\(a^3-b^3+ab\left(b-a\right)=\left(a-b\right)\left(a+b\right)^2\)
bien doi ve phai ta co:
\(\left(a-b\right)\left(a+b\right)^2\)
\(=a^3+ab^2-a^2b-b^3\)
\(=a^3-b^3+ab\left(b-a\right)\)= ve trai
vay \(a^3-b^3+ab\left(b-a\right)=\left(a-b\right)\left(a+b\right)^2\)
