Áp dụng BĐT Bu-nhi-a-cốp-ski,ta có :
\(\left(a^2+2\right)\left[1+\frac{\left(b+c\right)^2}{2}\right]\ge\left(a+b+c\right)^2\)
\(\Rightarrow\frac{1}{a^2+2}\le\frac{1+\frac{\left(b+c\right)^2}{2}}{\left(a+b+c\right)^2}\)
Tương tự : \(\frac{1}{b^2+2}\le\frac{1+\frac{\left(a+c\right)^2}{2}}{\left(a+b+c\right)^2}\) ; \(\frac{1}{c^2+2}\le\frac{1+\frac{\left(a+b\right)^2}{2}}{\left(a+b+c\right)^2}\)
Cộng vế theo vế,ta có :
\(\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\le\frac{3+\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2}{2}}{\left(a+b+c\right)^2}\)
\(=\frac{3+a^2+b^2+c^2+ab+bc+ac}{\left(a+b+c\right)^2}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu "=" xảy ra khi a = b = c = 1
Đặt \(P=\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\)
Thực hiện phép biến đổi theo biểu thức P ta được
\(Q=3-2P=\frac{a^2}{a^2+2}+\frac{b^2}{a^2+2}+\frac{c^2}{c^2+2}\)
Theo BĐT Cauchy-Schwarz ta có:
\(Q\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+6}=\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=1\)
\(\Rightarrow P\le1\). Dấu "=" xảy ra <=> a=b=c=1
tui chưa học bạn thông cảm
