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

\(\Leftrightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

b)

Ta có

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

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

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

\(\Leftrightarrow\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ac}{b+1}\le\frac{ab}{4\left(a+c\right)}+\frac{ab}{4\left(b+c\right)}+\frac{bc}{4\left(a+b\right)}+\frac{bc}{4\left(a+c\right)}+\frac{ac}{4\left(A+b\right)}+\frac{ac}{4\left(b+c\right)}\)

\(=\frac{ab+bc}{4\left(a+c\right)}+\frac{ab+ac}{4\left(b+c\right)}+\frac{bc+ac}{4\left(a+b\right)}=\frac{1}{4}\left(\frac{b\left(a+c\right)}{a+c}\right)+\frac{1}{4}\left(\frac{a\left(b+c\right)}{b+c}\right)+\frac{c\left(a+b\right)}{a+b}\)

\(=\frac{a+b+c}{4}=\frac{1}{4}\)

3+4+5

=7+5

=12

ai k mh mh k lại

bai nay kho day nha , cau cu lam theo nguyen ly di -rich -re la ra ngay dieu phai chung minh

Bất đẳng thức sai ngay khi a = b = \(\frac{1}{2}\)

Sai đề

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\varnothing\left(x^2\ge0\Rightarrow x^2+1\ge1>0\right)\end{matrix}\right.\)

Nghiệm của đa thức là x = 1

\(\Leftrightarrow\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)+\frac{1}{4}>0\)

\(\RightarrowĐPCM\)

Đk thiếu nhé : Đk : \(x\ge3;x\ne4\)

Có điều kiện \(x,y\in Z\) không ?

\(\frac{1}{4}\)là đáp án đúng.