本帖最后由 wanhuihua 于 2018-3-1 10:24 编辑
看看这样O不OK。
证明:令 $x=1/a$, $y=1/b$, $z=1/c$,则条件化为 $a$, $b$, $c>0$, $ab+bc+ca=3abc$,待 ...
kuing 发表于 2018-2-28 21:59
上面的证明实际上是调整法。(c+a)(c+b)>=(sqrt(ab) +c)^2 ,当a=b时成立。
$$
\eqalign{
& \cr
& {\cal 设}x,y,z{\cal 为}{\cal 正}{\cal 数}{\cal ,}{\cal 且}x{\text{ + }}y{\text{ + }}z{\text{ = 3}} \cr
& {\cal 求}{\cal 证}: \cr
& \frac{{\text{9}}}
{4}\, + \frac{1}
{{x^2 }} + \frac{1}
{{y^2 }} + \frac{1}
{{z^2 }} \geqslant \frac{7}
{4}(\,\frac{1}
{{xy}} + \,\frac{1}
{{yz}} + \,\frac{1}
{{zx}}) \cr
& {\cal 证}{\cal 明}{\cal :}{\cal 不}{\cal 妨}{\cal 设}x \leqslant y \leqslant z \cr
& {\cal 记}f{\text{(}}x{\text{,y) = }}\frac{{\text{9}}}
{4}\, + \frac{1}
{{x^2 }} + \frac{1}
{{y^2 }} + \frac{1}
{{z^2 }} - \frac{7}
{4}(\,\frac{1}
{{xy}} + \,\frac{1}
{{yz}} + \,\frac{1}
{{zx}}), \cr
& {\cal 则}t = f{\text{(}}x{\text{,y) - }}f{\text{(}}\frac{{x{\text{ + }}y}}
{2}{\text{,}}\frac{{x{\text{ + }}y}}
{2}{\text{) = }}\frac{1}
{{x^2 }} + \frac{1}
{{y^2 }} - \frac{8}
{{(x + y)^2 }} - \frac{7}
{4}(\,\frac{{x + y + z}}
{z})(\frac{1}
{{xy}} - \frac{4}
{{(x + y)^2 }}) \cr
& \geqslant \frac{1}
{{x^2 }} + \frac{1}
{{y^2 }} - \frac{8}
{{(x + y)^2 }} - \frac{{21}}
{4}(\frac{1}
{{xy}} - \frac{4}
{{(x + y)^2 }}) \cr
& t \geqslant 0 \Leftrightarrow \left( {4\,x^2 - 5\,xy + 4\,y^2 } \right)\left( {x - y} \right)^2 \geqslant 0,{\cal 显}{\cal 然}{\cal 成}{\cal 立} \cr
& {\cal 只}{\cal 需}{\cal 证}x{\text{ = }}y{\cal 时}{\cal :} \cr
& f{\text{ = }}\frac{{81}}
{{4{\cal (}x{\text{ + }}y{\text{ + }}z{\cal )}^2 }}\, + \frac{1}
{{x^2 }} + \frac{1}
{{y^2 }} + \frac{1}
{{z^2 }}{\text{ - }}\frac{7}
{4}(\,\frac{1}
{{xy}} + \,\frac{1}
{{yz}} + \,\frac{1}
{{zx}}) \geqslant 0{\cal 成}{\cal 立}{\cal 。} \cr
& f \geqslant 0 \Leftrightarrow (4x - z)^2 (x - z)^2 \geqslant 0{\cal 显}{\cal 然}{\cal 。} \cr
& {\text{Wanhuihua 20171210}} \cr
& \cr
& \cr}
$$ |