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Áp dụng BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\):
\(VT=\sqrt{\frac{x^2+\left(2y\right)^2}{2}}+\sqrt{\frac{\left(\frac{x}{2}-y\right)^2+3\left(\frac{x}{2}+y\right)^2}{3}}\)
\(VT\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{3\left(\frac{x}{2}+y\right)^2}{3}}\)
\(VT\ge\left|\frac{x+2y}{2}\right|+\left|\frac{x+2y}{2}\right|=\left|x+2y\right|\ge x+2y\) (đpcm)
Dấu "=" xảy ra khi \(x=2y\ge0\)
\(x^2+2y^2-2xy+2x-4y+2=0\)
\(\Rightarrow x^2-2xy+y^2+2\left(x-y\right)+1+y^2-2y+1=0\)
\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-1\right)^2=0\)
\(\Rightarrow\left(x-y+1\right)^2+\left(y-1\right)^2=0\)
=>................
đk: \(x+2y\ge0\)
\(x+2y=\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{\left(x+y\right)^2}{3}+y^2}\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{\left(x+2y\right)^2}{4}}=x+2y\)
\(\Rightarrow\)\(x=2y\)\(\Rightarrow\)\(x=3-y=3-\frac{x}{2}\)\(\Rightarrow\)\(\hept{\begin{cases}x=2\\y=\frac{x}{2}=1\end{cases}}\)
\(\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{x^2+2xy+4y^2}{3}}=\sqrt{\frac{x^2}{2}+\frac{4y^2}{2}}+\sqrt{\frac{\left(x+y\right)^2}{3}+\frac{y^2}{1}}\)
\(\ge\sqrt{\frac{\left(x+2y\right)^2}{2+2}}+\sqrt{\frac{\left(x+y+y\right)^2}{3+1}}=\frac{x+2y}{2}+\frac{x+2y}{2}=x+2y\)