\[ab+ bc+ ca= 3\]
\[\frac{a}{\left ( 1+ a \right )^{2}}+ \frac{b}{\left ( 1+ b \right )^{2}}+ \frac{c}{\left ( 1+ c \right )^{2}}\leq \frac{3}{4}\]
\[ab+ bc+ ca= 3\]
\[\frac{a}{\left ( 1+ a \right )^{2}}+ \frac{b}{\left ( 1+ b \right )^{2}}+ \frac{c}{\left ( 1+ c \right )^{2}}\leq \frac{3}{4}\]
Ta CM:
$\frac{a}{(1+a)^{2}}\leq \frac{1}{4}$ $\Leftrightarrow \frac{(a-1)^{2}}{4(a+1)^{2}}\geq 0$ (hiển nhiên đúng)
Tương tự suy ra:
$\frac{a}{(1+a)^{2}}+\frac{b}{(1+b)^{2}}+\frac{c}{(1+c)^{2}}\leq \frac{3}{4}$ (đpcm)
Dấu "=" xảy ra khi a=b=c=1
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