Question

[Hà Nam Phan Đình]

Đề: Cho \(\left\{{}\begin{matrix}x,y,z>0\\x+y\le z\end{matrix}\right.\) tìm Min của \(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\) Làm thế này không biết đúng ko

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

=> A \(=3+\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\right)\)

Áp dụng BĐT Cauchy ta có

\(A\ge3+2+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}\left(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\right)=6+\dfrac{15}{16}\left(\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}\right)\)

Do \(x+y\le z\Rightarrow\dfrac{x}{z}+\dfrac{y}{z}\le1\) ; Đặt \(u=\dfrac{x}{z}\); \(v=\dfrac{y}{z}\)

\(\Rightarrow\dfrac{z^2}{x^2}+\dfrac{z^2}{y^2}=\dfrac{1}{u^2}+\dfrac{1}{v^2}\ge\dfrac{2}{uv}\ge\dfrac{2}{\dfrac{\left(u+v\right)^2}{4}}\ge\dfrac{2}{\dfrac{1}{4}}=8\)

\(\Rightarrow A\ge6+\dfrac{15}{16}.8=\dfrac{27}{2}\) Vậy minA = \(\dfrac{27}{2}\) khi \(x=y=\dfrac{z}{2}\)

Answer 1

[Hà Nam Phan Đình]

@Unruly Kid

Answer 2

[Unruly Kid]

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