\(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\ge\frac{\left(a+b\right)^2}{a^2+b^2+2c^2}\)
\(\frac{b^2}{b^2+a^2}+\frac{c^2}{c^2+a^2}\ge\frac{\left(b+c\right)^2}{b^2+c^2+2a^2}\)
\(\frac{c^2}{c^2+b^2}+\frac{a^2}{a^2+b^2}\ge\frac{\left(c+a\right)^2}{c^2+a^2+2b^2}\)
\(\Rightarrow VT\le\frac{a^2+c^2}{a^2+c^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{a^2+b^2}{a^2+b^2}=1+1+1=3\)
Áp dụng BĐT Cauchy-Schwarz: \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Ta có \(\frac{\left(a+b\right)^2}{a^2+b^2+2c^2}=\frac{\left(a+b\right)^2}{a^2+c^2+b^2+c^2}\le\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\)
Tương tự ta có:
\(\frac{\left(b+c\right)^2}{b^2+c^2+2a^2}\le\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\) ; \(\frac{\left(c+a\right)^2}{c^2+a^2+2b^2}\le\frac{c^2}{b^2+c^2}+\frac{a^2}{a^2+b^2}\)
Cộng vế với vế:
\(\frac{\left(a+b\right)^2}{a^2+b^2+2c^2}+\frac{\left(b+c\right)^2}{b^2+c^2+2a^2}+\frac{\left(c+a\right)^2}{c^2+a^2+2b^2}\le\frac{a^2+c^2}{a^2+c^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{a^2+b^2}{a^2+b^2}=3\)
Dấu "=" xảy ra khi \(a=b=c\)
//Bạn chép đề sai, vế phải là số 3 chứ ko phải 1
