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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
\(\ge0\)\(\ge\)\(\ge\)
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:
\(\left(a^2+1\right)+\left(b^2+1\right)+\left(c^2+1\right)+\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\)
\(\ge2a+2b+2c+2ab+2bc+2ca=12\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)
\(\Rightarrow a^2+b^2+c^2\ge3\)
\(P=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}\)
\(P\ge a^2+b^2+c^2\ge3\)
\(P_{min}=3\) khi \(a=b=c=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)
\(\Rightarrow VT=\dfrac{1}{\dfrac{y}{x}\left(\dfrac{z}{y}+1\right)}+\dfrac{1}{\dfrac{z}{y}\left(\dfrac{x}{z}+1\right)}+\dfrac{1}{\dfrac{x}{z}\left(\dfrac{y}{x}+1\right)}\)
\(VT=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\dfrac{3}{2}\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\)
\(\Rightarrow x+y+z+xy+yz+zx=6\)
\(P=x^3+y^3+z^3\)
Ta có:
\(x^3+x^3+1\ge3x^2\)
Tương tự: \(2y^3+1\ge3y^2\) ; \(2z^3+1\ge3z^2\)
\(\Rightarrow2\left(x^3+y^3+z^3\right)\ge3\left(x^2+y^2+z^2\right)-3\)
\(\Rightarrow P\ge\dfrac{3}{2}\left(x^2+y^2+z^2-1\right)\)
Lại có: với mọi x;y;z thì:
\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)\ge2\left(x+y+z+xy+yz+zx\right)-3=9\)
\(\Rightarrow x^2+y^2+z^2\ge3\)
\(\Rightarrow P\ge\dfrac{3}{2}\left(3-1\right)=3\) (đpcm)