Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
So sánh a + b với 2c
Những câu hỏi liên quan
left(a+b+cright)^3-a^3-b^3-c^3left[left(a+bright)+cright]^3-a^3-b^3-c^3left(a+bright)^3+3left(a+bright)^2c+3left(a+bright)c^2+c^3-a^3-b^3-c^3a^3+3a^2b+3ab^2+b^3+3cleft(a^2+2ab+b^2right)+3ac^2+3bc^2-a^3-b^33a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^23left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abcright)3left[left(a^2b+ab^2right)+left(a^2c+abcright)+left(ac^2+bc^2right)+left(b^2c+abcright)right]3left[ableft(a+bright)+acleft(a+bright)+c^2left(a+bright)+bcleft(a+bright)right]3left(a+bright)left(ab+ac+c^2+bcrigh...
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\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)
\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)
\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)
Châu ơi!đăng làm j z
Cho a,b,c>0 và a+b+c=3Chứng minh \(\dfrac{a\left(a+bc\right)^2}{b\left(ab+2c^2\right)}+\dfrac{b\left(b+ca\right)^2}{c\left(bc+2a^2\right)}+\dfrac{c\left(c+ab\right)^2}{a\left(ca+2b^2\right)}\ge4\)
\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)}{a^2c^2+2ab^2c}\)
\(P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)
\(P\ge\dfrac{\left[a^2+b^2+c^2+3abc\right]^2}{\left(ab+bc+ca\right)^2}\)
Do đó ta chỉ cần chứng minh \(\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge2\)
Ta có: \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow3abc\ge4\left(ab+bc+ca\right)-9\)
\(\Rightarrow\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge\dfrac{a^2+b^2+c^2+4\left(ab+bc+ca\right)-9}{ab+bc+ca}\)
\(=\dfrac{\left(a+b+c\right)^2-9+2\left(ab+bc+ca\right)}{ab+bc+ca}=2\) (đpcm)
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sai cơ bản rồi bạn ơi : a(a+bc)^2 không bằng dc (a^2+abc)^2
a,b,c là các số thực dương thỏa mãn a+b+c=3. CMR: \(\dfrac{a\left(a+bc\right)^2}{b\left(ab+2c^2\right)}+\dfrac{b\left(b+ca\right)^2}{c\left(bc+2a^2\right)}+\dfrac{c\left(c+ab\right)^2}{a\left(ca+2b^2\right)}>=4\)
Trước hết theo BĐT Schur bậc 3 ta có:
\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)
Đặt vế trái BĐT cần chứng minh là P, ta có:
\(P=\dfrac{\left(a^2+abc\right)^2}{a^2b^2+2abc^2}+\dfrac{\left(b^2+abc\right)^2}{b^2c^2+2a^2bc}+\dfrac{\left(c^2+abc\right)^2}{a^2c^2+2ab^2c}\)
\(\Rightarrow P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)
Áp dụng (1):
\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
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Phân tích đa thức thành nhân tử:
a) \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
b) \(bc\left(b+c\right)+ac\left(c-a\right)-ab\left(a+b\right)\)
c) \(a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)+a^2c^2\left(c-a\right)\)
a: \(=a^2b-ab^2+b^2c-bc^2+c^2a-ca^2\)
\(=a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac+b^2-bc+c^2\right)\)
b: \(=b^2c+bc^2+ac^2-a^2c-a^2b-ab^2\)
\(=b^2\left(c-a\right)+b\left(c^2-a^2\right)+ac\left(c-a\right)\)
\(=\left(c-a\right)\left(b^2+ac+b\left(c+a\right)\right)\)
\(=\left(c-a\right)\left(b^2+ac+bc+ba\right)\)
\(=\left(c-a\right)\left(b+c\right)\left(b+a\right)\)
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Cho a,b,c >0 Chứng minh rằng :
\(\frac{c\left(a^2+b^2\right)^2}{b^3\left(ab+c^2\right)}+\frac{b\left(c^2+a^2\right)^2}{a^3\left(ac+b^2\right)}+\frac{a\left(b^2+c^2\right)^2}{c^3\left(bc+a^2\right)}\ge\frac{2\left(a^2b+b^2c+c^2a\right)}{abc}\)
mk mới hk lp 6 , bài này bó tay ko giải đc
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Có: a+b+c0Leftrightarrowleft(a+b+cright)^20Leftrightarrow a^2+b^2+c^2+2left(ab+bc+acright)0 Leftrightarrow2009+2left(ab+bc+acright)0 Leftrightarrow ab+bc+ac-frac{2009}{2} Leftrightarrowleft(ab+bc+acright)^2a^2b^2+b^2c^2+a^2c^2+2abcleft(a+b+cright)left(-frac{2009}{2}right)^2 Leftrightarrowleft(ab+bc+acright)^2a^2b^2+b^2c^2+a^2c^2left(-frac{2009}{2}right)^2 Mặt khác: left(a^2+b^2+c^2right)^2a^4+b^4+c^4+2left(a^2b^2+b^2c^2+c^2a^2right) a^4+b^4+c^4+2.left(-frac{2009}{2}right)^22009^2Leftrightarrow...
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Có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow2009+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-\frac{2009}{2}\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\left(-\frac{2009}{2}\right)^2\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2=\left(-\frac{2009}{2}\right)^2\)
Mặt khác: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=a^4+b^4+c^4+2.\left(-\frac{2009}{2}\right)^2=2009^2\)
\(\Leftrightarrow a^4+b^4+c^4=2009^2-2.\left(-\frac{2009}{2}\right)^2=2009^2-2.\frac{2009^2}{2^2}=2009^2-\frac{2009^2}{2}\)
--Hà Phương--
Cho \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Chứng minh rằng a=b=c
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac-4a^2-4b^2-4c^2+4ab+4bc+4ac=0\)
\(\Leftrightarrow-2a^2-2b^2-2c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow-\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)(đpcm)
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Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
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\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Cho a,b,c là độ dài 3 cạnh của 1 tam giác cm:
a)\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
b)\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)
c)\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\)
d)\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a+b\right)^2>a^3+b^3+c^3\)
a/ Với mọi số thực ta luôn có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Lại có do a;b;c là ba cạnh của 1 tam giác nên theo BĐT tam giác ta có:
\(a+b>c\Rightarrow ac+bc>c^2\)
\(a+c>b\Rightarrow ab+bc>b^2\)
\(b+c>a\Rightarrow ab+ac>a^2\)
Cộng vế với vế: \(2\left(ab+bc+ca\right)>a^2+b^2+c^2\)
b/
Do a;b;c là ba cạnh của tam giác nên các nhân tử vế phải đều dương
Ta có:
\(\left(a+b-c\right)\left(b+c-a\right)\le\frac{1}{4}\left(a+b-c+b+c-a\right)^2=b^2\)
Tương tự: \(\left(a+b-c\right)\left(a+c-b\right)\le a^2\)
\(\left(b+c-a\right)\left(a+c-b\right)\le c^2\)
Nhân vế với vế:
\(a^2b^2c^2\ge\left(a+b-c\right)^2\left(b+c-a\right)^2\left(a+c-b\right)^2\)
\(\Leftrightarrow abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)
\(VT=2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\)
\(=4a^2b^2-\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)\)
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
Mặt khác theo BĐT tam giác ta có:
\(\left\{{}\begin{matrix}a+b>c\\\left|a-b\right|< c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2>c^2\\\left(a-b\right)^2< c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-c^2>0\\c^2-\left(a-b\right)^2>0\end{matrix}\right.\)
\(\Rightarrow VT>0\)