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\(3,\)Áp dụng bđt Mincopski \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)hai lần có
\(VT\ge\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{yz}+\sqrt{zx}\right)^2}+\sqrt{z+xy}\)
\(\ge\sqrt{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}\)
\(=\sqrt{1+2t+t^2}\left(t=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\sqrt{\left(t+1\right)^2}=t+1=VP\left(Đpcm\right)\)
\(2,\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{a}.\sqrt{b}}}=\sqrt{\sqrt{ab}}\left(đpcm\right)\)
Áp dụng BĐT Cauchy , ta có :
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\ge\dfrac{x^3}{\dfrac{x^2+1-x^2}{2}}=2x^3\)
\(\dfrac{y^2}{\sqrt{1-y^2}}=\dfrac{y^3}{y\sqrt{1-y^2}}\ge\dfrac{y^3}{\dfrac{y^2+1-y^2}{2}}=2y^3\)
\(\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{z^3}{z\sqrt{1-z^2}}\ge\dfrac{z^3}{\dfrac{z^2+1-z^2}{2}}=2z^3\)
\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\)
\(0< x,y,z< 4\)\(\Rightarrow\)\(\hept{\begin{cases}x\left(x-4\right)< 0\\y\left(y-4\right)< 0\\z\left(z-4\right)< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2< 4x\\y^2< 4y\\z^2< 4z\end{cases}\Leftrightarrow}\hept{\begin{cases}x^3>\frac{x^4}{4}\\y^3>\frac{y^4}{4}\\z^3>\frac{z^4}{4}\end{cases}}}\)
\(\sqrt[4]{x^3}+\sqrt[4]{y^3}+\sqrt[4]{z^3}>\sqrt[4]{\frac{x^4}{4}}+\sqrt[4]{\frac{y^4}{4}}+\sqrt[4]{\frac{z^4}{4}}=\frac{x+y+z}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\)