Question

cho x và y là các số dương . CMR:  \(\frac{1}{x}\)+\(\frac{1}{y}\)>=\(\frac{4}{x+y}\)

Answer 1

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

<=>\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}\ge0\)

<=>\(\frac{y.\left(x+y\right)}{xy\left(x+y\right)}+\frac{x.\left(x+y\right)}{xy\left(x+y\right)}-\frac{4xy}{xy\left(x+y\right)}\ge0\)

<=>\(\frac{xy+y^2+x^2+xy-4xy}{xy\left(x+y\right)}\ge0\)

<=>\(\frac{x^2-2xy+y^2}{xy\left(x+y\right)}\ge0\)

<=>\(\frac{\left(x-y\right)^2}{xy\left(x+y\right)}\ge0\)đúng vơi mọi x;y dương ( vì (x-y)²\(\ge\)0)

vậy với x;y dương thì \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Answer 2

\(x^2+y^2\ge2xy\)\(\Rightarrow x^2+2xy+y^2\ge4xy\)\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\)\(\Rightarrow\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(đpcm)