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Áp dụng bất đẳng thức Co-si cho hai số không âm ta có:
\(x+y\ge2\sqrt{xy}\)
\(y+z\ge2\sqrt{yz}\)
\(z+x\ge2\sqrt{zx}\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8\sqrt{\left(xyz\right)^2}=8xyz\)
Dấu "=" <=> x = y = z. (đpcm)
Áp dụng BĐT cô-si cho 2 số dương ta có:
\(x+y\ge2\sqrt{xy}\)
\(y+z\ge2\sqrt{yz}\)
\(x+z\ge2\sqrt{xz}\)
=>\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{xz}=8\sqrt{x^2y^2z^2}=8xyz\)
Dấu"=" xảy ra <=>x=y y=z z=x=>x=y=z
=>\(\left(x+y\right)\left(y+z\right)\left(x+z\right)=8xyz\Leftrightarrow x=y=z\)(ĐPCM)
Áp dụng BĐT Cauchy cho 2 số không âm, ta được:
\(\frac{x+y}{2}\ge\sqrt{xy}\Rightarrow x+y\ge2\sqrt{xy}\)
\(\frac{y+z}{2}\ge\sqrt{yz}\Rightarrow y+z\ge2\sqrt{yz}\)
\(\frac{x+z}{2}\ge\sqrt{xz}\Rightarrow x+z\ge2\sqrt{xz}\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\sqrt{\left(xyz\right)^2}=8xyz\)(Vì x,y,z > 0)
Từ x+y+z=3 ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\frac{\Leftrightarrow xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)
Nhân chéo ta có:
\(\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)
\(\Leftrightarrow x^2y+xyz+x^2z+y^2x+y^2z+xyz+xyz+z^2y+z^2x=xyz\)
\(\Leftrightarrow x^2y+x^2z+y^2z+y^2x+z^2x+z^2y+2xyz=0\)
\(\Leftrightarrow\left(x^2y+x^2z+y^2x+xyz\right)+\left(y^2z+z^2x+z^2y+xyz\right)=0\)
\(\Leftrightarrow x\left(xy+xz+y^2+yz\right)+z\left(xy+xz+y^2+yz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=0\)
\(\Leftrightarrow\left(x+z\right)\left[\left(xy+y^2\right)+\left(xz+yz\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)
\(\Leftrightarrow\left(x+z\right)\left(y+z\right)\left(x+y\right)=0\)
Suy ra x+z=0 hoặc y+z=0 hoặc x+y=0
Với x+z=0 ta đc y=3
Với y+z=0 ta đc x=3
Với x+y=0 ta đc z=3
Từ đó suy ra đccm
Áp dụng BĐT Cosi cho 3 số x,y,z dương ta có:
\(x+y\ge2\sqrt{xy};y+z\ge2\sqrt{yz};z+x\ge2\sqrt{zx}\)
Nhân các BĐT vế theo vế ta được:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=8\sqrt{x^2y^2z^2}=8xyz\)
Dấu "=" xảy ra khi x = y = z
<=> x-y=y-z=z-x=0
<=>(x-y)2+(y-z)2+(z-x)2=0
<=>x2-2xy+y2+y2-2yz+z2+z2-2zx+x2=0
<=>2x2+2y2+2z2-2xy-2yz-2zx=0
<=>x2+y2+z2-xy-yz-zx=0
<=>(x+y+z)(x2+y2+z2-xy-yz-zx)=0 (vì x,y,z>0 nên x+y+z>0)
<=>x3+y3+z3-3xyz=0
<=>x3+y3+z3=3xyz (đpcm)
Áp dụng BĐT Cauchy cho 3 số dương, ta được:
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\)\(+\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\)
\(+\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}.3=\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(đpcm\right)\)
\(taco:\)
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge3\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{2}\ge3\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge3\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=\frac{3}{2}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(dpcm\right)\)
^^
Mình giải lại bài này cho đầy đủ hơn nhé: (nãy chỉ là hướng dẫn thôi)
Ta sẽ c/m: \(\frac{1}{x^2+x}\ge-\frac{3}{4}x+\frac{5}{4}\) (1).Thật vậy,xét hiệu hai vế,ta có:
\(VT-VP=\frac{\left(3x+4\right)\left(x-1\right)^2}{4\left(x^2+x\right)}\ge0\)
Suy ra \(VT\ge VP\).Vậy (1) đúng.
Thiết lập hai BĐT còn lại tương tự và cộng theo vế,ta có:
\(VT\ge-\frac{3}{4}\left(x+y+z\right)+\frac{5}{4}.3=\frac{3}{2}^{\left(đpcm\right)}\)