Có: a>1, b>1
=> a - 1> 0; b -1 >0
Áp dụng bđt Cauchy Schwarz dạng Engel có:
\(\dfrac{a^2}{b-1}+\dfrac{b^2}{a-1}\ge\dfrac{\left(a+b\right)^2}{\left(b-1+a-1\right)}=\dfrac{\left(a+b\right)^2}{\left(a+b-2\right)}\)
Ta cần cm: \(\dfrac{\left(a+b\right)^2}{\left(a+b-2\right)}\ge8\)
Có: \(\dfrac{\left(a+b\right)^2}{\left(a+b-2\right)}\ge8\)
\(\Leftrightarrow\left(a+b\right)^2\ge8\left(a+b\right)-16\)
\(\Leftrightarrow\left(a+b\right)^2-8\left(a+b\right)+16\ge0\)
\(\Leftrightarrow\left(a+b-4\right)^2\ge0\) (luôn đúng)
=> Đpcm
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}a=b\\a+b=4\end{matrix}\right.\)=> a = b = 2
Nay t rảnh nè :D
\(\dfrac{a^2}{b-1}+\dfrac{b^2}{a-1}\ge8\)
\(\Leftrightarrow\dfrac{a^2}{b-1}-4+\dfrac{b^2}{a-1}-4\ge0\)
\(\Leftrightarrow\dfrac{a^2-4b+4}{b-1}+\dfrac{b^2-4a+4}{a-1}\ge0\)
\(a-1;b-1>0\Leftrightarrow a^2-4b+4+b^2-4a+4\ge0\)
\(\Leftrightarrow\left(a-2\right)^2+\left(b-2\right)^2\ge0\) (đúng)
\("="\Leftrightarrow a=b=2\)
p/s: T ủng hộ cách mới,à ko,lm cách mới phá m cho vui