Nên bổ sung thêm đk a,b không âm

\(a+4b\ge\frac{16ab}{1+4ab}\)

\(\Leftrightarrow\left(a+4b\right)\left(1+4ab\right)\ge16ab\)

AM-GM:\(a+4b\ge4\sqrt{ab};1+4ab\ge4\sqrt{ab}\)

\(\Rightarrow\left(a+4b\right)\left(1+4ab\right)\ge16ab\left(đpcm\right)\)

Từ \(a+b=4ab\Leftrightarrow\frac{1}{a}+\frac{1}{b}=4\)

\(\left(\frac{1}{a};\frac{1}{b}\right)\rightarrow\left(x;y\right)\)\(\Rightarrow\hept{\begin{cases}x+y=4\\\frac{x^2}{4y+x^2y}+\frac{y^2}{4x+xy^2}\ge\frac{1}{2}\end{cases}}\)

C-S: \(VT\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)+xy\left(x+y\right)}\)\(\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)+\left(x+y\right)\cdot\frac{\left(x+y\right)^2}{4}}=\frac{1}{2}\)

vào tcn của tui ấn vào Thông kê hỏi đáp kéo xuống

là thế nào bạn ơi

\(a+b=4ab\le\left(a+b\right)^2\)

\(\frac{a}{4b^2+1}+\frac{b}{4a^2+1}=\frac{a^2}{4b^2a+a}+\frac{b^2}{4a^2b+b}\)

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

\("="\Leftrightarrow a=b=\frac{1}{2}\)

Cảm ơn bạn nhé.

\(a+b=4ab\Rightarrow\frac{1}{a}+\frac{1}{b}=4\Rightarrow4\ge\frac{4}{a+b}\Rightarrow a+b\ge1\)

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

\(=a-\frac{4ab^2}{4b^2+1}+b-\frac{4a^2b}{4a^2+1}\)

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

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

Ta có: \(\left(a+b\right)^2\ge4ab\Rightarrow-\frac{\left(a+b\right)^2}{2}\le-2ab\)

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

\("="\Leftrightarrow a=b=\frac{1}{2}\)

\(\dfrac{9}{4}=ab+a+b+1\le\dfrac{1}{4}\left(a+b\right)^2+a+b+1\)

\(\Leftrightarrow\left(a+b\right)^2+4\left(a+b\right)-5\ge0\)

\(\Leftrightarrow\left(a+b-1\right)\left(a+b+5\right)\ge0\)

\(\Leftrightarrow a+b-1\ge0\) (do \(a+b+5>0\))

\(\Rightarrow a+b\ge1\)

b.

\(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\ge\dfrac{1}{2}.1^2=\dfrac{1}{2}\) (đpcm)

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

\(\ge a-\dfrac{4ab^2}{4b}+b-\dfrac{4a^2b}{4a}\) (bđt Cô-si)

=a-ab+b-ab=a+b-2ab=4ab-2ab=2ab

Lại có a+b=4ab \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

Dấu ''='' xảy ra khi \(a=b=\dfrac{1}{2}\)

\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

\(\Leftrightarrow a-\dfrac{a}{4b^2+1}+b-\dfrac{b}{4a^2+1}\le a+b-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{4ab^2}{4b^2+1}+\dfrac{4ba^2}{4a^2+1}\le4ab-\dfrac{1}{2}\)

\(\sum\dfrac{4ab^2}{4b^2+1}\le^{CS}2ab\)

\(\Rightarrow CM:2ab\le4ab-\dfrac{1}{2}\Leftrightarrow ab\ge\dfrac{1}{4}\)

Từ GT \(\Rightarrow4ab=a+b\ge2\sqrt{ab}\Leftrightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow dpcm\)