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2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)
\(P^2=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2.\left(\frac{xy.yz}{zx}+\frac{yz.zx}{xy}+\frac{zx.xy}{zy}\right)\)
\(=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2.2016\)
Áp dụng BĐT Cauchy:\(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}\ge2\sqrt{\frac{x^2y^2}{z^2}.\frac{y^2z^2}{x^2}}=2y^2\)
\(\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge2\sqrt{\frac{y^2z^2}{x^2}.\frac{z^2x^2}{y^2}}=2z^2\)
\(\frac{z^2x^2}{y^2}+\frac{x^2y^2}{z^2}\ge2\sqrt{\frac{x^2z^2}{y^2}.\frac{x^2y^2}{z^2}}=2x^2\)
Cộng theo vế ta được:\(2\left(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\right)\ge2x^2+2y^2+2z^2=2.2016\)
\(\Rightarrow\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge2016\)
\(\Rightarrow P^2\ge2016+2016.2=6048\Rightarrow P\ge\sqrt{6048}=12\sqrt{42}\)
Nên GTNN của P là \(12\sqrt{42}\) đạt được khi \(x=y=z=\sqrt{\frac{2016}{3}}=4\sqrt{42}\)
C/m: \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
\(\Rightarrow2x^2+xy+2y^2\ge\dfrac{5}{4}\left(x^2+2xy+y^2\right)\)
\(\Leftrightarrow8x^2+4xy+8y^2\ge5x^2+10xy+5y^2\)
\(\Leftrightarrow3\left(x-y\right)^2\ge0\left(LĐ\right)\)
Vậy \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
CMTT: \(\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)\);
\(\sqrt{2z^2+zx+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)
Vậy H=\(\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+xz+2z^2}\ge\sqrt{5}\left(x+y+z\right)=2019\)Hmin=2019\(\Leftrightarrow x=y=z=\dfrac{\dfrac{2019}{\sqrt{5}}}{3}\)
Lời giải:
\(P=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\)
Xét
\((x^2y^2+y^2z^2+z^2x^2)^2=x^4y^4+y^4z^4+z^4x^4+2x^2y^2z^2(x^2+y^2+z^2)\) (1)
Áp dụng BĐT AM-GM:
\(x^4y^4+y^4z^4\geq 2x^2y^4z^2\)
\(y^4z^4+z^4x^4\geq 2x^2y^2z^4\)
\(x^4y^4+z^4x^4\geq 2x^4y^2z^2\)
Cộng theo vế: \(\Rightarrow 2(x^4y^4+y^4z^4+z^4x^4)\geq 2x^2y^2z^2(x^2+y^2+z^2)\)
\(\Leftrightarrow x^4y^4+y^4z^4+z^4x^4\geq x^2y^2z^2(x^2+y^2+z^2)\) (2)
Từ \((1);(2)\Rightarrow (x^2y^2+y^2z^2+z^2x^2)^2\geq 3x^2y^2z^2(x^2+y^2+z^2)\)
\(\Leftrightarrow (x^2y^2+y^2z^2+z^2x^2)^2\geq 6048x^2y^2z^2\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\geq 12\sqrt{42}xyz\)
\(\Leftrightarrow P=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\geq 12\sqrt{42}\)
Vậy \(P_{\min}=12\sqrt{42}\Leftrightarrow x=y=z=4\sqrt{42}\)