Áp dụng BĐT Bunhia cho 2 cặp số (\(\sqrt{x}\) ;y); (\(\sqrt{x}\);1)
Ta có: ( \(\sqrt{x}\)\(\sqrt{x}\) +1y )2 \(\le\) (\(\sqrt{x}\)2 +1)(\(\sqrt{x}\)2 +y2)
\(\Leftrightarrow\) (x + y )2 \(\le\) (x + 1)( x + y2)
\(\Leftrightarrow\) x + y2 \(\ge\) \(\dfrac{\left(x+y\right)^2}{x+1}\) ( Do x + 1> 0 )
Áp dụng ta có: a+b2 \(\ge\) \(\dfrac{\left(a+b\right)^2}{a+1}\)
\(\Leftrightarrow\) \(\dfrac{1}{a+b^2}\le\dfrac{a+1}{\left(a+b\right)^2}\)
CMTT: \(\dfrac{1}{b+a^2}\le\dfrac{b+1}{\left(a+b\right)^2}\)
\(\Rightarrow\) M \(\le\) \(\dfrac{a+b+2}{\left(a+b\right)^2}\)
Xét hiệu \(\dfrac{a+b+2}{\left(a+b\right)^2}\) -1
= \(\dfrac{a+b+2-\left(a+b\right)^2}{\left(a+b\right)^2}\) = \(\dfrac{4\left(a+b\right)+8-4\left(a+b\right)^2}{4\left(a+b\right)^2}\)
= \(\dfrac{-\left[2\left(a+b\right)-1\right]^2+9}{4\left(a+b\right)^2}\)
Do a + b \(\ge\) 2 \(\Rightarrow\) \(\dfrac{a+b+2}{\left(a+b\right)^2}\) -1 \(\le\) 0
\(\Rightarrow\) \(\dfrac{a+b+2}{\left(a+b\right)^2}\) \(\le\)1
\(\Rightarrow\) M \(\le\)1
Vậy Mmax = 1
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=1
