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VT=\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy.\left(x+y+z\right)\)
\(=\left(x+y\right)^2-\left(x+y\right).z+z^2-3xy\left(\text{vì }x+y+z=1\right)\)
\(=x^2+2xy+y^2-xz-yz+z^3-3xy\)
\(=x^2+y^2+z^2-xy-yz-xz\)
\(=\frac{1}{2}.\left(2x^2+2y^2+2z^2-2xy-2yz-2xz\right)\)
\(=\frac{1}{2}.\left[\left(x^2-2xy-y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\right]\)
\(=\frac{1}{2}.\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)=VP
=>dpcm
Ta có : \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=x+y+z\left(x^2+y^2+z^2+2xy+xz+yz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=x^2+y^2+z^2-xy-yz-xz=\frac{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)}{2}=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)
Câu 2:
Từ điều kiện bài này có thể đặt ẩn phụ và AM-GM ra luôn kết quả, nhưng hơi rắc rối khi người ta hỏi từ đâu mà có cách đặt ẩn phụ như vậy, do đó ta giải trâu :D
\(x^2+y^2+z^2+xyz=4\)
\(\Leftrightarrow\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{4}+2\left(\frac{x}{2}.\frac{y}{z}.\frac{z}{2}\right)=1\)
\(\Leftrightarrow\frac{xy}{2z}.\frac{xz}{2y}+\frac{xy}{2z}.\frac{yz}{2x}+\frac{yz}{2x}.\frac{xz}{2y}+2\left(\frac{xy}{2z}.\frac{yz}{2x}.\frac{xy}{2y}\right)=1\)
Đặt \(\left(\frac{xy}{2z};\frac{zx}{2y};\frac{yz}{2x}\right)=\left(m;n;p\right)\Rightarrow mn+np+pn+2mnp=1\)
\(\Leftrightarrow2\left(n+1\right)\left(m+1\right)\left(p+1\right)=\left(n+1\right)\left(m+1\right)+\left(n+1\right)\left(p+1\right)+\left(m+1\right)\left(p+1\right)\)
\(\Leftrightarrow\frac{1}{n+1}+\frac{1}{m+1}+\frac{1}{p+1}=2\)
\(\Leftrightarrow1=\frac{n}{n+1}+\frac{m}{m+1}+\frac{p}{p+1}\ge\frac{\left(\sqrt{n}+\sqrt{m}+\sqrt{p}\right)^2}{m+n+p+3}\)
\(\Leftrightarrow m+m+p+2\left(\sqrt{mn}+\sqrt{np}+\sqrt{mp}\right)\le m+n+p+3\)
\(\Leftrightarrow\sqrt{mn}+\sqrt{np}+\sqrt{mp}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\le\frac{3}{2}\Leftrightarrow x+y+z\le3\)
Câu 1:
\(2xyz=1-\left(x+y+z\right)+xy+yz+zx\)
\(\Rightarrow xy+yz+zx=2xyz+\left(x+y+z\right)-1\)
\(VT=x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\)
\(=\left(x+y+z\right)^2-2\left(x+y+z\right)-4xyz+2\)
\(VT\ge\left(x+y+z\right)^2-2\left(x+y+z\right)-\frac{4}{27}\left(x+y+z\right)^3+2\)
\(VT\ge\frac{4}{27}\left[\frac{15}{4}-\left(x+y+z\right)\right]\left(x+y+z-\frac{3}{2}\right)^2+\frac{3}{2}\ge\frac{3}{2}\)
(Do \(0< x;y;z< 1\Rightarrow x+y+z< 3< \frac{15}{4}\))
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
Khó wá bạn ơi mk chịu
Mk mới chỉ học lớp 5 thôi
Ai đồng ý thì
Arigatouuuuuuuuuuuuu
Ta có : \(\left(x+y+z\right)\left(x^2+y^2+z^2\right)\le3\left(x^3+y^3+z^3\right)\)
\(\Leftrightarrow2\left(x^3+y^3+z^3\right)-x^2\left(y+z\right)-y^2\left(x+z\right)-z^2\left(x+y\right)\ge0\)
\(\Leftrightarrow x^2\left(x-y\right)+x^2\left(x-z\right)+y^2\left(y-x\right)+y^2\left(y-z\right)+z^2\left(z-x\right)+z^2\left(z-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)+\left(y-z\right)\left(y^2-z^2\right)+\left(z-x\right)\left(z^2-x^2\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)+\left(y-z\right)^2\left(y+z\right)+\left(z-x\right)^2\left(z+x\right)\ge0\) (luôn đúng vì x,y,z > 0)
Vậy bđt ban đầu được chứng minh
\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)
=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)
=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)
Dấu = xảy ra khi x=y=z=6căn 2
Dễ lắm bạn! Biến đổi tương đương là ok á!
\(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(VT\ge3\left[\frac{\left(x+y+z\right)^2}{3}\right]=\frac{3\left(x+y+z\right)^2}{3}=VP\left(đpcm\right)\)(bất đẳng thức svacxo)