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Ta co: \(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=y+z\)
Thê vào ta được
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
Áp dụng bđt AM-GM ta có
\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)
Ta có \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng bđt AM-GM ta có
\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)
Làm tiếp bài ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ chớ hình như bị ngược dấu ó.Do mình gà nên chỉ biết cô si mù mịt thôi ạ
\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)
\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)
\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)
\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
Bài này thì AM-GM thôi
\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Sử dụng BĐT AM-GM cho 3 số không âm ta có :
\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)^2}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)
\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)
\(=3\sqrt[3]{\left(\frac{xy}{x}+\frac{1}{x}\right)\left(\frac{yz}{y}+\frac{1}{y}\right)\left(\frac{zx}{z}+\frac{1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Tiếp tục sử dụng AM-GM cho 2 số không âm ta được :
\(3\sqrt[3]{\left(2\sqrt[2]{y\frac{1}{x}}\right)\left(2\sqrt[2]{z\frac{1}{y}}\right)\left(2\sqrt[2]{x\frac{1}{z}}\right)}\ge3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right)\left(2\sqrt{\frac{z}{y}}\right)\left(2\sqrt{\frac{x}{z}}\right)}\)
\(=3\sqrt[3]{8\left(\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}\right)}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)
Vậy \(Min_P=6\)đạt được khi \(x=y=z=\frac{1}{2}\)
\(P=\dfrac{6}{2xy+2yz+2zx}+\dfrac{2}{x^2+y^2+z^2}\ge\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=8+4\sqrt{3}\)
Thay \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)
\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)
Xét hạng tử: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\)
Thay \(xy+yz+zx=1\); ta có:
\(x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)^2\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}=xy+xz\)
Tượng tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=xy+yz;z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=xz+yz\)
Do đó: \(A=2\left(xy+yz+zx\right)=2.1=2\)
ĐS:...
Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)
Theo giả thiết,ta có: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}=\frac{3}{abc}\)
Nhân hai vế với abc: \(a+b+c=3\) tức là \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Lại có:\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{xyz}\)
Ta cần c/m: \(A\ge\frac{3}{2}\)
Do x,y,z > 0 áp dụng BĐT Cô si: \(x^3+y^3+z^3\ge3xyz=xy+yz+zx\)
Áp dụng BĐT Cô si: \(A\ge3\sqrt[3]{\frac{x^3y^3z^3}{\left(z+x^2\right)\left(x+y^2\right)\left(y+z^2\right)}}\)
\(=3xyz.\frac{1}{\sqrt[3]{\left(z+x^2\right)\left(x+y^2\right)\left(y+z^2\right)}}\)\(\ge3xyz.\frac{xy+yz+zx}{\left(x+y+z\right)+\left(x^2+y^2+z^2\right)}\)
\(=\frac{3\left(x^2y^2z+xy^2z^2+x^2yz^2\right)}{\left(x+y+z\right)+\left(x^2+y^2+z^2\right)}\ge\frac{3x^2y^2z^2}{\left(x+y+z\right)+\left(x^2+y^2+z^2\right)}\)
\(=\frac{3x^2y^2z^2}{\left(x+y+z\right)+\left(x+y+z\right)^2-2\left(xy+yz+zx\right)}\)
\(=\frac{3x^2y^2z^2}{\left(x+y+z\right)\left(x+y+z+1\right)-6xyz}\)
\(=\frac{3x^2y^2z^2}{xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left[xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+1\right]-6xyz}\)
\(=\frac{3x^2y^2z^2}{3xyz\left[3xyz+1\right]-6xyz}=\frac{3x^2y^2z^2}{9x^2y^2z^2-3xyz}\)
Đặt \(B=\frac{1}{A}=\frac{9x^2y^2z^2-3xyz}{3x^2y^2z^2}\)
Ta sẽ c/m: \(B\ge\frac{2}{3}\).Thật vậy,ta có:
\(B=\frac{1}{A}=\frac{9x^2y^2z^2-3xyz}{3x^2y^2z^2}=3-\frac{3}{3xyz}\)\(=3-\frac{1}{xyz}\ge0\)
Suy ra \(A\ge0?!?\) có gì đó sai sai.Ai biết chỉ giùm
Nghĩ mãi mới ra -.- Để ý cái số mũ 3 trên tử khó mà dùng trực tiếp Cô-si hoặc Bunhia nên phải tách nó ra
Ta có: \(\frac{x^3}{x^2+z}=\frac{x^3+xz}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\)
\(\ge x-\frac{xz}{2x\sqrt{z}}\)(Cô-si)
\(=x-\frac{\sqrt{z}}{2}\)
\(\ge x-\frac{z+1}{4}\)(Dùng bđt \(\sqrt{z}\le\frac{z+1}{2}\))
Tương tự \(\frac{y^3}{y^2+z}\ge y-\frac{x+1}{4}\)
\(\frac{z^3}{z^2+y}\ge z-\frac{y+1}{4}\)
Cộng từng vế của các bđt trên lại được
\(A\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{4}\)
\(=\frac{3\left(x+y+z\right)}{4}-\frac{3}{4}\)
Từ điều kiện \(xy+yz+zx=3xyz\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Áp dụng bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(a,b,c>0\right)\)được
\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Rightarrow x+y+z\ge3\)
Quay trở lại với A
\(A\ge\frac{3\left(x+y+z\right)}{4}-\frac{3}{4}\ge\frac{3.3}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)(Do \(3=\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\))
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y=z\\xy+yz+zx=3\end{cases}\Leftrightarrow x=y=z=1}\)
Vậy .............
Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)
\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)
\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\)
(vì \(2013=3.671=3\left(xy+yz+zx\right)\))
\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)
\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)
\(=\dfrac{1}{x+y+z}\)
ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)
\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)
\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))
Vậy ta có đpcm.
