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áp dụng bđt cauchy:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{x}.\frac{1}{y}}=\frac{2}{\sqrt{xy}}.\)
Tượng tự \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}},\frac{1}{z}+\frac{1}{x}\ge\frac{2}{\sqrt{xz}}.\)
=>2VT>=2Vp
<=>VT>=VP
dấu = xảy ra khi x=y=z
By AM-GM we have:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}};\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}};\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{xz}}\)
Cộng theo vế rồi rút gọn là có ĐPCM
Xảy ra khi x=y=z
#)Giải :
Ta có : \(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\left(1\right)\)
\(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\left(2\right)\)
\(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{xz}}\left(3\right)\)
Cộng (1),(2),(3) vế theo vế ta được :
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\left(đpcm\right)\)
Ta thấy : \(\left(x-y\right)^2\ge0\)\(\Rightarrow x^2+y^2\ge2xy\)
Mà : \(x^2+y^2=1\)\(\Rightarrow2xy\le1\)
\(\Rightarrow x^2+y^2+2xy\le1+1\)
\(\Rightarrow\left(x+y\right)^2\le2\)
\(\Leftrightarrow|x+y|\le\sqrt{2}\)
\(\Rightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)\(\left(đpcm\right)\)
\(P=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+1}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{xz}+\sqrt{z}+1}\)( Vì xyz=1 nên \(\sqrt{xyz}=1\))
\(P=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{y}+1+\sqrt{yz}\right)}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{z}\left(\sqrt{x}+1+\sqrt{xy}\right)}\)
\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{1}{\sqrt{x}+1+\sqrt{xy}}\)
\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{\sqrt{xyz}}{\sqrt{x}\left(1+\sqrt{yz}+\sqrt{y}\right)}\)
\(P=\frac{\sqrt{y}+1}{\sqrt{y}+1+\sqrt{yz}}+\frac{\sqrt{yz}}{\sqrt{y}+1+\sqrt{yz}}=\frac{\sqrt{y}+1+\sqrt{yz}}{\sqrt{y}+1+\sqrt{yz}}=1\)
tiếp tục câu 2,vì máy bị lỗi nên phải tách ra:
Ta có:\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3\left(xy+xz+yz\right)\right).\)
Dó đó:\(x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\left(x+y+z\right)^2-3\left(xy+yz+xz\right)+2010\right)\)
\(=\left(x+y+z\right)^3.\)(2)
TỪ \(\left(1\right),\left(2\right)\)suy ra \(P\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}.\)
Dấu \(=\)xảy ra khi \(x=y=z=\frac{\sqrt{2010}}{3}\)
2)Ta có:
\(x\left(x^2-yz+2010\right)=x\left(x^2+xy+xz+1340\right)>0\)
Tương tự ta có:\(y\left(y^2-xz+2010\right)>0,z\left(z^2-xy+2010\right)>0\)
Áp dụng svac-xơ ta có:
\(P=\frac{x^2}{x\left(x^2-yz+2010\right)}+\frac{y^2}{y\left(y^2-xz+2010\right)}+\frac{z^2}{z\left(z^2-xy+2010\right)}\)
\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}.\)(1)
Đặt \(\hept{\begin{cases}\sqrt{x}=p\\\sqrt{y}=q\\\sqrt{z}=r\end{cases}}\). Khi đó \(\hept{\begin{cases}p+q+r=1\\p,q,r>0\end{cases}}\)
và ta cần chứng minh \(\frac{pq}{\sqrt{p^2+q^2+2r^2}}+\frac{qr}{\sqrt{q^2+r^2+2p^2}}+\frac{rp}{\sqrt{r^2+p^2+2q^2}}\le\frac{1}{2}\)
Ta có: \(\frac{pq}{\sqrt{p^2+q^2+2r^2}}=\frac{2pq}{\sqrt{\left(1+1+2\right)\left(p^2+q^2+2r^2\right)}}\)
\(\le\frac{2pq}{p+q+2r}\le\frac{1}{2}\left(\frac{pq}{p+r}+\frac{pq}{q+r}\right)\)(Theo BĐT Cauchy-Schwarz và BĐT \(\frac{1}{u}+\frac{1}{v}\ge\frac{4}{u+v}\)) (1)
Hoàn toàn tương tự: \(\frac{qr}{\sqrt{q^2+r^2+2p^2}}\le\frac{1}{2}\left(\frac{qr}{q+p}+\frac{qr}{r+p}\right)\)(2); \(\frac{rp}{\sqrt{r^2+p^2+2q^2}}\le\frac{1}{2}\left(\frac{rp}{r+q}+\frac{rp}{p+q}\right)\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{pq}{\sqrt{p^2+q^2+2r^2}}+\frac{qr}{\sqrt{q^2+r^2+2p^2}}+\frac{rp}{\sqrt{r^2+p^2+2q^2}}\)\(\le\frac{1}{2}\left(\frac{r\left(p+q\right)}{p+q}+\frac{p\left(q+r\right)}{q+r}+\frac{q\left(r+p\right)}{r+p}\right)=\frac{1}{2}\left(p+q+r\right)=\frac{1}{2}\)(Do p + q + r = 1)
Đẳng thức xảy ra khi \(p=q=r=\frac{1}{3}\)hay \(x=y=z=\frac{1}{9}\)
\(\frac{1}{\sqrt{xy}}\)<= {\(\frac{1}{x}\)+\(\frac{1}{y}\)} : 2
Tương tư.....
=> DPCM