Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\frac{2a}{a+b}\cdot\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}\cdot\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}\cdot\frac{c}{2\left(b+c\right)}}\)
\(\le\frac{1}{2}\left(\frac{2a}{a+b}+\frac{2b}{a+b}+\frac{2a}{a+c}+\frac{2c}{a+c}+\frac{b}{2\left(b+c\right)}+\frac{c}{2\left(b+c\right)}\right)\)
\(=\frac{1}{2}\left(2+2+\frac{1}{2}\right)=\frac{9}{4}\)
Áp dụng BĐT Cauchy-Schwarz ta có :
\(P=\frac{2a}{\sqrt{1+a^2}}+\frac{b}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}\)
\(=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\frac{2a}{a+b}.\frac{2a}{a+c}}+\sqrt{\frac{2b}{a+b}.\frac{b}{2\left(b+c\right)}}+\sqrt{\frac{2c}{a+c}.\frac{c}{2\left(b +c\right)}}\)
\(\le\frac{1}{2}\left(\frac{2a}{a+b}+\frac{2b}{a+b}+\frac{2a}{a+c}+\frac{2c}{a+c}+\frac{b}{2\left(b+c\right)}+\frac{c}{2\left(b+c\right)}\right)\)
\(=\frac{1}{2}\left(2+2+\frac{1}{2}\right)=\frac{9}{4}\)
P/s : Mình tự nghĩ chứ không phải mình copy đâu
P/s: Ok @Nguyen Huy Hoang mình biết bạn ko phải thằng copy đâu :)
