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Lời giải:
Áp dụng BĐT AM-GM:
$1=xy+yz+xz+2xyz\leq \frac{(x+y+z)^2}{3}+2.\frac{(x+y+z)^3}{27}$
$\Leftrightarrow 1\leq \frac{t^2}{3}+\frac{2t^3}{27}$ (đặt $x+y+z=t$)
$\Leftrightarrow 2t^3+9t^2-27\geq 0$
$\Leftrightarrow (t+3)^2(2t-3)\geq 0$
$\Leftrightarrow 2t-3\geq 0$
$\Leftrightarrow t\geq \frac{3}{2}$ hay $x+y+z\geq \frac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $x=y=z=\frac{1}{2}$
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.
Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)
\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2}\)
\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Dấu "=" <=> x=y=z=1
