choa,b,c>0 ab+bc+ca=1
tìm maxP=$\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}+\frac{3c}{\sqrt{1+c^{2}}}$
choa,b,c>0 ab+bc+ca=1
tìm maxP=$\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}+\frac{3c}{\sqrt{1+c^{2}}}$
Ta có : $\frac{a}{1+a^2}+\frac{b}{1+b^2}=\frac{1}{a+b}\left ( \frac{a}{a+c}+\frac{b}{b+c} \right )=\frac{ab+1}{(a+b)\left ( 1+c^2 \right )}$
Lại có: $ab+bc+ca=1\Rightarrow ab+1=2-c(a+b) \Rightarrow \frac{ab+1}{a+b}=\frac{2}{a+b}-c$
$c^{2}+1=(c+a)(c+b)\leq \frac{(a+b+2c)^2}{4}\Rightarrow a+b\geq 2\left ( \sqrt{c^2+1}-c \right )$
Từ đó ta có : $\frac{ab+1}{(a+b)(1+c^2)}\leq \frac{1}{1+c^2}\left ( \frac{1}{\sqrt{c^2+1} -c}-c \right )=\frac{1}{\sqrt{1+c^2}}$
$\Rightarrow P\leq \frac{1+3c}{\sqrt{1+c^2}}=\sqrt{10}-\frac{\left ( c-3 \right )^2}{\sqrt{c^2+1}}\leq \sqrt{10}$
Đẳng thức xảy ra $\Leftrightarrow a=b=\sqrt{10}-3,c=3$
0 thành viên, 0 khách, 0 thành viên ẩn danh