Question

a, Cho x , y \(\ge\)1 . CMR : \(\dfrac{1}{1+x^2}\) + \(\dfrac{1}{1+y^2}\)\(\ge\)\(\dfrac{2}{1+xy}\)

b, Cho x \(\ge\)1 , y\(\ge\)0 và 6xy +2x - 3y \(\le\) 2 . Tìm GTNN

A = \(\dfrac{1}{4x^2-4x+2}\)+ \(\dfrac{1}{9y^2+6y+2}\)

Answer 1

a/ Cho x, y ≥ 1. Chứng minh: 1/(1 + x^2) + 1/(1 + y^2) ≥ 2/(1 + xy)

b/ Đề:...Tìm GTLN

Có:

\(\dfrac{1}{4x^2-4x+2}=\dfrac{1}{\left(2x-1\right)^2+1}\le\dfrac{1}{2}\forall x\ge1\)

\(\dfrac{1}{9y^2+6y+2}=\dfrac{1}{\left(3y+1\right)^2+1}\le\dfrac{1}{2}\forall y\ge0\)

\(\Rightarrow A=\dfrac{1}{4x^2-4x+2}+\dfrac{1}{9y^2+6y+2}\le\dfrac{1}{2}+\dfrac{1}{2}=1\)

Vậy MAX_A = 1 khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)