Question

Cho a , b > 0 và x+y =4

C/m : \(\dfrac{x^2}{a}+\dfrac{y^2}{b}\ge\dfrac{16}{a+b}\)

Answer 1

Áp dụng BĐT Cauchy - Schwarz:

\(\dfrac{x^2}{a}+\dfrac{y^2}{b}\ge\dfrac{\left(x+y\right)^2}{a+b}=\dfrac{4^2}{a+b}=\dfrac{16}{a+b}\)

\("="\Leftrightarrow x=y=2\)

Answer 2

Áp dụng BĐT Cauchy dạng Engel , ta có :

\(\dfrac{x^2}{a}+\dfrac{y^2}{b}\ge\dfrac{\left(x+y\right)^2}{a+b}=\dfrac{16}{a+b}\)
\("="\Leftrightarrow\left\{{}\begin{matrix}x=y=2\\a=b\end{matrix}\right.\)

Answer 3

Áp dụng BĐT Cauchy dạng Engel , ta có :

\(\dfrac{x^2}{a}+\dfrac{y^2}{b}\ge\dfrac{\left(x+y\right)^2}{a+b}=\dfrac{16}{a+b}\)

\("="\Leftrightarrow x=y=2\)