Có x+y=1 \Rightarrow xy\leq\frac{1}{4} Có x2 + \frac{1}{y2}=x2 + \frac{1}{16y2}+\frac{1}{16y2}+...+\frac{1}{16y2}(16 số \frac{1}{16y2})\geq 17\sqrt[17]{\frac{x2}{(16y2)16}}Cmtt: y2 + \frac{1}{x2} \geq 17\sqrt[17]{\frac{y2}{(16x2)16}}\Rightarrow P \geq 17\sqrt[17]{\frac{x2}{(16y2)16}}*17\sqrt[17]{\frac{y2}{(16x2)16}}=172 *\sqrt[17]{\frac{1}{1617 *(16x2y2)15}}\geq 172 *\frac{1}{16}=\frac{289}{16}dấu = có khi x=y=\frac{1}{2}
Có
$x+y=1 \Rightarrow xy\leq\frac{1}{4}
$Có
$x
^2 + \frac{1}{y
^2}
$$=x
^2 + \frac{1}{16y
^2}+\frac{1}{16y
^2}+...+\frac{1}{16y
^2}(16 số \frac{1}{16y
^2})\geq 17\sqrt[17]{\frac{x2}{(16y
^2)
^{16}}
}$Cmtt:
$y
^2 + \frac{1}{x
^2} \geq 17\sqrt[17]{\frac{y
^2}{(16x
^2)
^{16}}
}$$\Rightarrow P \geq 17\sqrt[17]{\frac{x
^2}{(16y
^2)16}}*17\sqrt[17]{\frac{y
^2}{(16x
^2)16}}=17
^2 *\sqrt[17]{\frac{1}{1617 *(16x
^2y
^2)15}}\geq 17
^2 *\frac{1}{16}=\frac{289}{16}
$dấu = có khi
$x=y=\frac{1}{2}
$