Question

Answer 1

Ta có: \(xy+yz+zx\le\dfrac{1}{3}\left(x+y+z\right)^2=3\)

\(\Rightarrow A\le\dfrac{xy}{\sqrt{z^2+xy+yz+zx}}+\dfrac{yz}{\sqrt{x^2+xy+yz+zx}}+\dfrac{zx}{\sqrt{y^2+xy+yz+zx}}\)

\(\Rightarrow A\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}+\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt{\left(x+y\right)\left(y+z\right)}}\)

\(A\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}+\dfrac{yz}{x+y}+\dfrac{yz}{x+z}+\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)=\dfrac{x+y+z}{2}=\dfrac{3}{2}\)

\(A_{max}=\dfrac{3}{2}\) khi \(x=y=z=1\)

Answer 2

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