Tìm max $Q=2xy+yz+zx$

V

vansang02121998

Áp dụng AM-GM:

$(2-\sqrt{3})x^2+\dfrac{1}{2}z^2 \ge (\sqrt{3}-1)xz$

$(2-\sqrt{3})y^2+\dfrac{1}{2}z^2 \ge (\sqrt{3}-1)yz$

$(\sqrt{3}-1)x^2+(\sqrt{3}-1)y^2 \ge 2(\sqrt{3}-1)xy$

$\Rightarrow x^2+y^2+z^2 \ge (\sqrt{3}-1)(2xy+yz+xz)$

$\Rightarrow 2xy+xz+yz \le \dfrac{1}{\sqrt{3}-1} = \dfrac{\sqrt{3}+1}{2}$

Dấu $"="$ xảy ra khi $x=y=\sqrt[4]{\dfrac{2+\sqrt{3}}{24}};z=\sqrt[4]{\dfrac{2-\sqrt{3}}{6}}$
 
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