Question

chứng minh rằng a^2/b + c^2/d >=(a+c)^2/(b+d)

Answer 1

Áp dụng bđt bunhia copski, ta có

\(\left(a+c\right)^2=\left(\dfrac{a\sqrt{b}}{\sqrt{b}}+\dfrac{c\sqrt{d}}{\sqrt{d}}\right)^2\le\left[\left(\dfrac{a}{\sqrt{b}}\right)^2+\left(\dfrac{c}{\sqrt{d}}\right)^2\right]\left[\left(\sqrt{b}\right)^2+\left(\sqrt{d}\right)^2\right]=\left(\dfrac{a^2}{b}+\dfrac{c^2}{d}\right)\left(b+d\right)\Leftrightarrow\left(a+c\right)^2\le\left(\dfrac{a^2}{b}+\dfrac{c^2}{d}\right)\left(b+d\right)\Leftrightarrow\dfrac{a^2}{b}+\dfrac{c^2}{d}\ge\dfrac{\left(a+c\right)^2}{b+d}\)

Dấu bằng xảy ra khi \(\dfrac{a}{b}=\dfrac{c}{d}\)