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yeumtnguoithtk


ta có :4xy \leq (x + y)^2
\Leftrightarrow4xy\leq4
\Leftrightarrow xy = 1
tuong tu ta cung co
4(2xy)(x^2+y^2)\leq(2xy+(x^2+y^2))^2\Leftrightarrow8xy(x^2+y^2)\leq(x+y)^4
\Leftrightarrow8xy(x^2+y^2)\leq16\Leftrightarrowxy(x^2 + y^2)\leq2
bay gio toi su dung 2 ten nhan vat nha
1la yeumtnguoithk va congtu200860 nhe nhung chu yeu la congtu200860
\Leftrightarrow4xy\leq4
\Leftrightarrow xy = 1
tuong tu ta cung co
4(2xy)(x^2+y^2)\leq(2xy+(x^2+y^2))^2\Leftrightarrow8xy(x^2+y^2)\leq(x+y)^4
\Leftrightarrow8xy(x^2+y^2)\leq16\Leftrightarrowxy(x^2 + y^2)\leq2
bay gio toi su dung 2 ten nhan vat nha
1la yeumtnguoithk va congtu200860 nhe nhung chu yeu la congtu200860
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