Với mọi x;y dương, ta có:
\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow2x^2+2y^2\ge x^2+y^2+2xy\)
\(\Leftrightarrow x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\)
Đồng thời \(x^2+y^2\ge2xy\Rightarrow x^2+y^2+2xy\ge4xy\Rightarrow\left(x+y\right)^2\ge4xy\)
\(\Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Áp dụng: đặt vế trái của BĐT cần chứng minh là P, ta có:
\(P=\left(a+\dfrac{1}{b}\right)^2+\left(b+\dfrac{1}{a}\right)^2\ge\dfrac{1}{2}\left(a+\dfrac{1}{b}+b+\dfrac{1}{a}\right)^2=\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)
\(P\ge\dfrac{1}{2}\left(a+b+\dfrac{4}{a+b}\right)^2=\dfrac{1}{2}\left(3+\dfrac{4}{3}\right)^2=\dfrac{169}{18}\)
Dấu "=" xảy ra khi \(a=b=\dfrac{3}{2}\)
