Question

cho x;y>0 va x+y<=2 tim min cua

\(M=\frac{20}{x^2+y^{^2}}+\frac{11}{xy}\)

 

Answer 1

\(M=\frac{20}{x^2+y^2}+\frac{11}{xy}=\frac{20}{x^2+y^2}+\frac{22}{2xy}=\frac{20}{x^2+y^2}+\frac{20}{2xy}+\frac{2}{2xy}\)

\(=20\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{xy}>=20\cdot\frac{4}{x^2+2xy+y^2}+\frac{4}{\left(x+y\right)^2}\)

\(=\frac{80}{\left(x+y\right)^2}+\frac{4}{\left(x+y\right)^2}=\frac{84}{\left(x+y\right)^2}>=\frac{84}{2^2}=\frac{84}{4}=21\)

dấu = xảy ra khi \(\hept{\begin{cases}x+y=2\\x=y\end{cases}\Rightarrow x=y=1}\)

vậy min M là 21 khi x=y=1