CMR: c(a^2+b^2) + a(b^2+c^2) + b(c^2+a^2) >= 6abc
Những câu hỏi liên quan
Cho a, b , c > 0
CMR: a^2(b+c) +b(a^2+c^2) +c(a^2+b^2) >=6abc
đpcm\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\left(lđ\right)\)
Dấu bằng xảy ra khi a=b=c
Cho a,b,c dương CMR: a(b2+c2) +b(a2+c2)+c(a2+b2) => 6abc
a^2+b^2>= 2ab thì a(b^2+c^2) >= 2abc . Làm tương tự suy ra dpcm
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Cho a,b,c thỏa mãn (a - b)2 + (b - c)2 + (c - a)2 = 6abc
CMR: a3 + b3 + c3 = 3abc(a + b + c +1)
Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=6abc\)
\(\Leftrightarrow a^2+b^2+c^2-\left(ab+bc+ac\right)=3abc\)
\(\Leftrightarrow\left(a+b+c\right)^2-3\left(ab+bc+ac\right)=3abc\)
Đặt \(\left(a+b+c,ab+bc+ac,abc\right)=\left(p,q,r\right)\)
\(\Rightarrow p^2-3q=3r\)
Khi đó \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Leftrightarrow a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ac\right)+3abc\)
\(\Leftrightarrow a^3+b^3+c^3=p^3+3pq+3r=p\left(p^2-3q\right)+3r=3pr+3r\)
Vậy .....
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chờ a,b ,c thỏa mãn
a+b+c+ab+bc+ca=6abc
cmr:\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\supseteq3\)
Cho a,c,b dương t/m a+b+c+ab+bc+ac = 6abc
CMR : \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Đặt \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)
CMTT : \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc};\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2.}{ca}\)
Ta có : \(\left(\frac{1}{a}-1\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+1\ge\frac{2}{a}\)
CMTT : \(\frac{1}{b^2}+1\ge\frac{2}{b};\frac{1}{c^2}+1\ge\frac{2}{c}\)
\(3A+3\ge2.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2.6=12\)
\(\Leftrightarrow A+1\ge4\Leftrightarrow A\ge3\left(đpcm\right)\)
Chúc bạn học tốt !!!
\(a+b+c+ab+ac+bc=6abc\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Đặt \(\hept{\begin{cases}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{cases}}\) \(\Rightarrow x+y+z+xy+xz+yz=6\)
Cần chứng minh \(P=x^2+y^2+z^2\ge3\)
Ta có BĐT quen thuộc :
\(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
\(2x^2+2y^2+2z^2\ge2xy+2xz+2yz\)
Cộng vế với vế :
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)\)
Dấu " = " xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
Từ \(a+b+c+ab+bc+ac=6\left(1\right)\)
Vì a,b,c dương nên ta chia hai vế của pt (1) cho abc ta có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=6\)
Ta có
\(\frac{1}{a^2}+1\ge\frac{2}{a}\)
\(\frac{1}{b^2}+1\ge\frac{2}{b}\)
\(\frac{1}{c^2}+1\ge\frac{2}{c}\)
Và
\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\)
\(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)
Công theo BĐT ta có
\(3\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge2\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow3\cdot\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge12\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\ge4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu \("="\)xảy ra \(\Leftrightarrow a+b+c=1\)
cho a,b,c > 0 thỏa mãn a+b+c+ab+bc+ca=6abc
Cmr: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Cho 3 số a, b, c. Biết \(a+b+c+ab+bc+ca=6abc\). CMR: \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)
Ta có a,b,c dương⇒\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\dfrac{1}{cb}+\dfrac{1}{ac}+\dfrac{1}{ab}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}=6\)(1)
Đặt x=\(\dfrac{1}{a}\),y=\(\dfrac{1}{b}\),z=\(\dfrac{1}{c}\)
Vậy (1)\(\Leftrightarrow xy+xz+yz+x+y+z=6\)
Áp dụng bđt cosi ta có
\(x^2+1\ge2x\)(2)
\(y^2+1\ge2y\)(3)
\(z^2+1\ge2z\)(4)
Cộng (2),(3),(4)\(\Leftrightarrow x^2+y^2+z^2+3\ge2x+2y+2z\)(5)
Ta lại có bất đẳng thức cosi:
\(x^2+y^2\ge2xy\)(6)
\(y^2+z^2\ge2yz\)(7)
\(x^2+z^2\ge2xz\)(8)
Cộng (6),(7),(8)\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2xy+2xz+2yz\left(9\right)\)
Cộng (8),(9)\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2.6\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge9\Leftrightarrow x^2+y^2+z^2\ge3\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\Rightarrowđpcm\)
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Với a,b,c >0 và a+b+c+ab+bc+ca=6abc. CMR: 1/a2+1/b2+1/c2 lớn hơn hoặc bằng 3
Cho a, b, c>0 thoả mãn a+b+c+ab+bc+ca=6abc
cmr:\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
\(GT\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Ta có:
\(2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Cộng vế với vế:
\(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+3\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
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