Question

cho x, y dương, chứng minh \(\dfrac{x}{y^2}+\dfrac{y}{x^2}+\dfrac{16}{x+y}\ge5\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

Answer 1

Đặt \(\left(x;y\right)=\left(\dfrac{1}{a};\dfrac{1}{b}\right)\)

BĐT trở thành: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}+\dfrac{16ab}{a+b}\ge5\left(a+b\right)\)

\(\Leftrightarrow\dfrac{a^3+b^3}{ab}+\dfrac{16ab}{a+b}-5\left(a+b\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a^3+b^3\right)+16a^2b^2-5ab\left(a+b\right)^2}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^4}{ab\left(a+b\right)}\ge0\) (luôn đúng)