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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)\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+xz+yz\right)\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge2.6\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge9\Leftrightarrow x^2+y^2+z^2\ge3\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\Rightarrowđpcm\)
a, b, c khác 0 nhé
\(a+b+c+ab+bc+ca=6abcd\)
Chia cả hai vế cho abc ta có
\(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=6\)
Đặt \(\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\), x, y, z khác 0
bài toán đưa về cho 3 số x, y, z khác 0 chứng minh x+y+z+xy+yz+xz=6 Chứng minh rằng x^2+y^2+z^2>=3
Xét 3(x^2+y^2+z^2)- 2(x+y+z+xy+xz+yz) +3=(x^2-2xy+y^2)+(x^2-2xz+z^2)+(z^2-2zy+y^2)+(x^2-2x+1)+(y^2-2y+1)+(z^2-2z+1)
=(x-y)^2+(x-z)^2+(z-y)^2+(x-1)^2+(y-1)^2+(z-1)^2\(\ge\)0
=> 3(x^2+y^2+z^2)- 2(x+y+z+xy+xz+yz) +3\(\ge0\)=> 3.(x^2+y^2+z^2)-2.6+3\(\ge0\)<=> x^2+y^2+z^2\(\ge\)3 (điều phải chứng minh)
Dấu '=" xảy ra khi và chỉ khi x=y=z=1
\(\ge0\)\(\ge\)\(\ge\)
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 !!!
\(a+b+c+ab+bc+ca=6abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
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 \(a=b=c=1\)
Lời giải:
Áp dụng BĐT AM-GM (Cô-si)
\(1+a^3+b^3\geq 3\sqrt[3]{a^3b^3}=3ab\)
\(\Rightarrow \frac{\sqrt{1+a^3+b^3}}{ab}\geq \frac{\sqrt{3ab}}{ab}=\frac{c\sqrt{3ab}}{abc}=c\sqrt{3ab}=\sqrt{c}.\sqrt{3abc}=\sqrt{3c}\)
Hoàn toàn tương tự:
\(\frac{\sqrt{1+b^3+c^3}}{bc}\geq \sqrt{3a}\)
\(\frac{\sqrt{1+a^3+c^3}}{ac}\geq \sqrt{3b}\)
Cộng theo vế những BĐT vừa thu được ta có:
\(\frac{\sqrt{a^3+b^3+1}}{ab}+\frac{\sqrt{b^3+c^3+1}}{bc}+\frac{\sqrt{c^3+a^3+1}}{ac}\geq \sqrt{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})\)
\(\geq \sqrt{3}.3\sqrt[3]{\sqrt{a}.\sqrt{b}.\sqrt{c}}=\sqrt{3}.3\sqrt[6]{abc}=3\sqrt{3}\) (áp dụng BĐT Cô-si)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
Ta có : \(3\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\Rightarrow a+b+c\ge3\)
Theo BĐT AM-GM ta có :
\(\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)
Tương tự :
\(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2}\)
\(\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)
\(\Rightarrow\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2}+\dfrac{1}{2}\left(ab+bc+ca\right)\ge\left(a+b+c\right)-\dfrac{1}{2}\left(ab+bc+ca\right)+\dfrac{1}{2}\left(ab+bc+ca\right)=a+b+c\ge3\)
a+b+c+ab+bc+ca=6abc \(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)
Đặt \(A=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
Ta có: \(\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2\ge0\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab}\)
CMTT: \(\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ca}\)
Ta có: \(\left(\dfrac{1}{a}-1\right)^2\ge0\Leftrightarrow\dfrac{1}{a^2}+1\ge\dfrac{2}{a}\)
CMTT: \(\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)
\(3A+3\ge2.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=2.6=12\)
<=> A + 1 \(\ge4\Leftrightarrow A\ge3\) (đpcm)
con súc vật đừng có tag tao vào tao đéo thích giúp loại như mày