a. Từ giả thiết ta có:
\(\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow x^2+y^2=4-2xy\ge4-2.\frac{\left(x+y\right)^2}{4}=4-2.\frac{4}{4}=2\)
\(\Rightarrow Min=2\Leftrightarrow x=y=1\)
b. Từ giả thiết suy ra:
\(3\ge\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca\le1\)
\(\Rightarrow T=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\)
\(\le\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{a}{\sqrt{\left(c+b\right)\left(a+b\right)}}+\frac{a}{\sqrt{\left(c+b\right)\left(a+c\right)}}\)
\(=\sqrt{\frac{a}{a+b}.\frac{a}{a+c}}+\sqrt{\frac{b}{c+b}.\frac{b}{a+b}}+\sqrt{\frac{a}{b+c}.\frac{a}{a+c}}\)
\(\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{c+b}+\frac{b}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}\right)\)
\(=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{1}{2}\left(1+1+1\right)=\frac{3}{2}\)
\(Max_T=\frac{3}{2}\Leftrightarrow a=b=c=\frac{\sqrt{3}}{3}\)
