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a+b+c+ab+bc+ac = 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 !!!
\(P=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\) ; \(Q=\frac{1}{2}\left(ab+ac+bc\right)\)
\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{1}{2}ab\)
Tương tự và cộng lại: \(P\ge a+b+c-Q\Rightarrow P+Q\ge a+b+c\)
Mặt khác \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Rightarrow a+b+c\ge\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{9}{3}=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\)\(\Rightarrow x+y+z+xy+yz+zx=6\)
CM \(P=x^2+y^2+z^2\ge3\)
\(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
Cộng vế với vế
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge3\left(đpcm\right)\)
Vậy dấu "=" xảy ra khi \(x=y=z=1\) hoặc \(a=b=c=1\)
\(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Ta lại có:
\(\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1-3\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}-3\)
\(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\ge\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
Cộng vế với vế:
\(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)-3\)
\(\Leftrightarrow3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge6.2-3=9\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
From \(a+b+c+ab+bc+ca=6abc\)
\(\Rightarrow\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
Let \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow\left(x;y;z\right)\) we have
\(x^2+y^2+z^2\ge3\forall\hept{\begin{cases}x+y+z+xy+yz+xz=6\\x,y,z>0\end{cases}}\)
Áp dụng BĐT AM-GM ta có:
\(x^2+1\ge2\sqrt{x^2}=2x\)
\(y^2+1\ge2\sqrt{y^2}=2y\)
\(z^2+1\ge2\sqrt{z^2}=2z\)
Cộng theo vế 3 BĐT trên ta có:
\(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\left(1\right)\)
Lại có BĐT quen thuộc \(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\left(2\right)\)
Cộng theo vế của \(\left(1\right);\left(2\right)\) ta có:
\(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\cdot6=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\Rightarrow x^2+y^2+z^2\ge3\)
Đẳng thức xảy ra khi \(a=b=c=1\)
#Nguồn:Xem câu hỏi (tui tự chép tui hihi :v)
P = \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}\)
hay 2P \(\ge\frac{2\left(a+b+c\right)}{abc}\) (1)
mặt khác theo Cauchy ta có \(\frac{1}{a^2}+1\ge\frac{2}{a}\)
do đó P \(\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3\) hay P \(\ge\frac{2\left(ab+bc+ca\right)}{abc}-3\) (2)
từ (1) và (2) suy ra 3P \(\ge\frac{2\left(a+b+c+ab+bc+ca\right)}{abc}-3=9\)
hay P \(\ge\)3
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(a+b+c+ab+ac+bc=6abc\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Hay \(x+y+z+xy+yz+xz=6\)
Cần chứng minh \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=x^2+y^2+z^2\ge3\)
Ta có : \(\left(x^2+1\right)+\left(y^2+1\right)+\left(z^2+1\right)\ge2\left(x+y+z\right)\) (BĐT Cosi)
\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\) (BĐT Cosi)
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)=12\)
\(\Rightarrow x^2+y^2+z^2\ge3\) (đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)
\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)
Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{a^2}}\)
Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.
Từ dk suy ra 1/bc+1/ac+1/ab+1/c+1/b+1/a=6 đặt 1/a=x;1/b=y;1/c=z→x+y+x+xy+yz+xz=6 ta phải cm x2+y2+z2>=3 Ta có:2(x2+y2+z2)>=2(xy+yz+xz) (1) (x-1)2>=0→x2>=2x-1 Tương tự :y2>=2y-1;z2>=2z-1 do đó :x2+y2+z2>=2(x+y+z)-3 (2) cộng vế 1 vs 2 ta có:3(x2+y2+z2)>=2(x+y+z+xy+yz+xz)-3 <=>3(x2+y2+z2)>=2.6-3 <=>x2+y2+z2>=3
\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
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 \(a=b=c=1\)