Question

bài 1:Cho a>0;b>0 thỏa mãn a+b=1

CMR \(\frac{1}{ab}+\frac{1}{a^2+b^2}\ge6\)

Answer 1

\(VT=\left(\frac{1}{2ab}+\frac{1}{a^2+b^2}\right)+\frac{1}{2ab}\)

\(\ge\frac{4}{\left(a+b\right)^2}+\frac{1}{2ab}=4+\frac{1}{2ab}\)

Ta có: \(\frac{\left(a+b\right)^2}{4}\ge ab\Rightarrow\frac{\left(a+b\right)^2}{2}\ge2ab\) (BĐT AM-GM or CÔ si gì đó)

\(VT\ge4+\frac{1}{\frac{\left(a+b\right)^2}{2}}=4+2=6^{\left(đpcm\right)}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a^2+b^2=2ab\\a+b=1\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\a+b=1\end{cases}}\Leftrightarrow}\hept{\begin{cases}a=b\\a+b=1\end{cases}}\Leftrightarrow a=b=\frac{1}{2}\)