Áp dụng BĐT Cô-si , ta có :

\(a+b\ge2\sqrt{ab}\)    và     \(\frac{1}{a}+\frac{1}{b}\ge\frac{1}{\sqrt{ab}}\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Rightarrow\frac{1}{a+b}=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(đpcm)

Dấu " = " xảy ra \(\Leftrightarrow a=b\)

Dòng thứ 5 nhầm dấu ạ :D Sửa :

\(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)           

(đpcm)

\(ĐKXĐ:x\ge0\)

Ta có: \(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}=\frac{x-1}{x+2\sqrt{x}+1}\)

\(\Rightarrow A^2=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}=\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}\)

\(\Rightarrow A^2+A=\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}+\frac{x-1}{x+2\sqrt{x}+1}\)

\(=\frac{2x-2\sqrt{x}}{x+2\sqrt{x}+1}=\frac{2\left(x-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)^2}\)

\(A\le0\Leftrightarrow\orbr{\begin{cases}A=0\\A< 0\end{cases}}\)

+) A = 0\(\Leftrightarrow2\left(x-\sqrt{x}\right)=0\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\left(tm\right)\)

+) A < 0 \(\Leftrightarrow2\left(x-\sqrt{x}\right)< 0\)(vì \(\left(\sqrt{x}+1\right)^2>0\forall x\ge0\)

\(\Leftrightarrow x-\sqrt{x}< 0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}\\\sqrt{x}-1\end{cases}}\)trái dấu

Mà \(\sqrt{x}>\sqrt{x}-1\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\\sqrt{x}< 1\end{cases}}\Leftrightarrow0< x< 1\)

Vậy 0 < x < 1 thì \(A^2+A\le0\)

Sửa)): 

\(0\le x\le1\)nha. Ghi nhầm dấu ở kết luận

Do 2 th là \(\hept{\begin{cases}x=0;x=1\\0< x< 1\end{cases}}\Rightarrow\)\(0\le x\le1\)