[toán 9] tìm min

C

congchuaanhsang

Ta có $\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0$ \Leftrightarrow $z=\dfrac{2xy}{x+y}$

Do đó $x+z=\dfrac{x^2+3xy}{x+y}$ ; $2x-z=\dfrac{2x^2}{x+y}$

$y+z=\dfrac{y^2+3xy}{x+y}$ ; $2y-z=\dfrac{2y^2}{x+y}$

Nên $T=\dfrac{x^2+3xy}{2x^2}+\dfrac{y^2+3xy}{2y^2}=1+(\dfrac{3x}{2y}+\dfrac{3y}{2x})$

\geq$1+3=4$ (Cauchy)

Dấu "=" bạn tự tìm
 
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