Áp dụng BĐT Côsi ta có:
\(P=\left(a+\frac{1}{b}+1\right)^2+\left(b+\frac{1}{a}+1\right)^2\ge\frac{\left(a+\frac{1}{b}+1+b+\frac{1}{a}+1\right)^2}{2}\) (BĐT quen thuộc)
\(=\frac{1}{2}\left[\left(\frac{1}{a}+\frac{4}{361}a\right)+\left(\frac{1}{b}+\frac{4}{361}b\right)+\frac{357}{361}\left(a+b\right)+2\right]^2\)
\(\ge\frac{1}{2}\left(\frac{4}{19}+\frac{4}{19}+\frac{357}{361}\cdot19+2\right)^2=\left(\frac{403}{38}\right)^2\)
Dấu "='' xảy ra khi: \(a=b=\frac{19}{2}\)
Sai thì bỏ qua:))
\(\left(a+\frac{1}{b}+1\right)^2+\left(b+\frac{1}{a}+1\right)^2\ge\frac{\left[\left(a+\frac{1}{b}+1\right)+\left(b+\frac{1}{a}+1\right)\right]^2}{2}\)\(=\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}+2\right)^2}{2}\)
\(\ge\frac{\left(a+b+\frac{4}{a+b}+2\right)^2}{2}=\frac{\left(19+\frac{4}{19}+2\right)^2}{2}=...\)
Dấu đẳng thức xảy ra khi \(a=b=\frac{19}{2}\)
\(P=\left(x+\frac{1}{y}+1\right)^2+\left(y+\frac{1}{x}+1\right)^2\)
\(2P=\left[\left(x+\frac{1}{y}+1\right)^2+\left(y+\frac{1}{x}+1\right)^2\right]\left(1^2+1^2\right)\ge\left(x+\frac{1}{y}+y+\frac{1}{x}+2\right)^2\)
\(=\left(21+\frac{1}{x}+\frac{1}{y}\right)^2\ge\left(21+\frac{4}{x+y}\right)^2=\left(\frac{403}{19}\right)^2\)
Suy ra \(P\ge\frac{1}{2}\left(\frac{403}{19}\right)^2\)
Dấu \(=\)xảy ra khi \(x=y=\frac{19}{2}\).
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