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theo bđt cauchy schwarz ta có
\(\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2\sqrt{y^3z^2}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2\sqrt{z^3y^2}}=\dfrac{1}{zy}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}+\dfrac{\dfrac{1}{y^2}+\dfrac{1}{z^2}}{2}+\dfrac{\dfrac{1}{z^2}+\dfrac{1}{x^2}}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)\(\Rightarrow dpcm\)
\(P=\sqrt{\dfrac{x^3}{y+3}}+\sqrt{\dfrac{y^3}{z+3}}+\sqrt{\dfrac{z^3}{x+3}}\)
\(=\dfrac{x^2}{\sqrt{x\left(y+3\right)}}+\dfrac{y^2}{\sqrt{y\left(z+3\right)}}+\dfrac{z^2}{\sqrt{z\left(x+3\right)}}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}}\)
Xét:
\(\left(\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\right)^2\le\left(1^2+1^2+1^2\right)\left(xy+3x+yz+3y+xz+3z\right)\)
\(=3\left(9+xy+yz+xz\right)\)
\(=27+3\left(xy+yz+xz\right)\le27+\left(x+y+z\right)^2=36\)
\(\Rightarrow\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\le6\)
\(P\ge\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=1\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\ge\dfrac{x^3}{\dfrac{x^2+1-x^2}{2}}=2x^3\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Cộng theo vế 3 BĐT trên ta có:
\(P\ge2x^3+2y^3+2z^3=2\left(x^3+y^3+z^3\right)=2\)
\(\dfrac{x}{x+\sqrt{x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}\)\(\ge\dfrac{x}{x+\sqrt{xz}+\sqrt{xy}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Đặt vế trái là P, ta có:
\(P\le\sqrt{3\left(\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\right)}\)
Nên ta chỉ cần chứng mình: \(\sqrt{3\left(\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\right)}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3x}{z+3x}-1+\dfrac{3y}{x+3y}-\dfrac{3z}{y+3z}-1\le\dfrac{9}{4}-3\)
\(\Leftrightarrow\dfrac{z}{z+3x}+\dfrac{x}{x+3y}+\dfrac{y}{y+3z}\ge\dfrac{3}{4}\)
BĐT trên đúng do:
\(\dfrac{z}{z+3x}+\dfrac{x}{x+3y}+\dfrac{y}{y+3z}=\dfrac{z^2}{z^2+3zx}+\dfrac{x^2}{x^2+3xy}+\dfrac{y^2}{y^2+3yz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\dfrac{1}{3}\left(x+y+z\right)^2}=\dfrac{3}{4}\)