\(P=\frac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{3\left(a+b\right)}{4\sqrt{ab}}+\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)

\(\Rightarrow P\ge\frac{3.2\sqrt{ab}}{4\sqrt{ab}}+2\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}=\frac{3}{2}+1=\frac{5}{2}\)

\(\Rightarrow P_{min}=\frac{5}{2}\) khi a=b

Lời giải:

\(a+b=ab\Rightarrow \frac{1}{a}+\frac{1}{b}=1\)

Đặt \(\left(\frac{1}{a}, \frac{1}{b}\right)=(x,y)\) thì bài toán trở thành:

Cho $x,y>0$ thỏa mãn $x+y=1$. Tìm GTNN của biểu thức:

\(P=\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{\sqrt{(x^2+1)(y^2+1)}}{xy}\)

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Áp dụng BĐT Cauchy-Schwarz, AM-GM:

\(\frac{x^2}{2x+1}+\frac{y^2}{2y+1}\geq \frac{(x+y)^2}{2x+1+2y+1}=\frac{1}{2+2}=\frac{1}{4}\)

\((x^2+1)(y^2+1)\geq (xy+1)^2\Rightarrow \frac{\sqrt{(x^2+1)(y^2+1)}}{xy}\geq \frac{xy+1}{xy}=1+\frac{1}{xy}\)

\(\geq 1+\frac{1}{\frac{(x+y)^2}{4}}=5\)

\(\Rightarrow P=\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{\sqrt{(x^2+1)(y^2+1)}}{xy}\geq \frac{1}{4}+5=\frac{21}{4}\)

Vậy \(P_{\min}=\frac{21}{4}\Leftrightarrow x=y=\frac{1}{2}\Leftrightarrow a=b=2\)

p \(\ge\)\(\frac{4}{a^2+b^2+2\left(a+b\right)}\) +\(\sqrt{\left(1+ab\right)^2}\) (bunhia và cosi)

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

do \(a+b=ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow a+b\ge4\)

dạt a+b = t thì t>=4

cần tìm min \(\frac{4}{t^2}+t+1=\frac{4}{t^2}+\frac{t}{16}+\frac{t}{16}+\frac{7t}{8}+1\)

                                      \(\ge3.\sqrt[3]{\frac{4}{t^2}.\frac{t}{16}.\frac{t}{16}}+\frac{7.4}{8}+1=\frac{21}{4}\)

dau = xay ra khi a=b=2

\(\sqrt{ab}+\sqrt{4b.c}+2\left(a+c\right)\le\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(4b+c\right)+2\left(a+c\right)=\dfrac{5}{2}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{2}{5}\left(\dfrac{1}{a+b+c}-\dfrac{1}{\sqrt{a+b+c}}\right)=\dfrac{2}{5}\left(\dfrac{1}{\sqrt{a+b+c}}-\dfrac{1}{2}\right)^2-\dfrac{1}{10}\ge-\dfrac{1}{10}\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a+b+c=4\\a=b=\dfrac{c}{4}\end{matrix}\right.\) em tự giải ra a;b;c

\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

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

Tương tự: \(\dfrac{1}{\left(1+c\right)^2}+\dfrac{1}{\left(1+d\right)^2}\ge\dfrac{1}{1+cd}\)

\(\Rightarrow B\ge\dfrac{1}{1+ab}+\dfrac{1}{1+cd}=\dfrac{1}{1+ab}+\dfrac{1}{1+\dfrac{1}{ab}}=\dfrac{1}{1+ab}+\dfrac{ab}{1+ab}=1\)

\(B_{min}=1\) khi \(a=b=c=d=1\)

Áp dụng BĐT phụ ta có:

\(B\ge\dfrac{1}{1+ab}+\dfrac{1}{1+cd}=\dfrac{ab+cd+2}{1+ab+cd+abcd}=1\)

Vậy GTNN của B bằng 1 <=> a=b=c=d=1

\(P=\frac{3\left(a+b\right)}{\sqrt{9a\left(4a+5b\right)}+\sqrt{9b\left(4b+5a\right)}}\)

\(\ge\frac{3\left(a+b\right)}{\frac{9a+4a+5b}{2}+\frac{9b+4b+5a}{2}}=\frac{1}{3}\)

Ta có :

  \(P^1=\frac{a+b}{\sqrt{a\left(4a+5b\right)}+\sqrt{b\left(4b+5a\right)}}.\)

\(\Leftrightarrow P^2=\frac{3\left(a+b\right)}{\sqrt{9a\left(4a+5b\right)}+\sqrt{9b\left(4b+5a\right)}}\)

Mà ta thấy  biểu thức \(P^2\ge\frac{3\left(a+b\right)}{\frac{9a+4a+5b}{2}+\frac{9b+4b+5a}{2}}\)

                                     \(=\frac{1}{3}\)

Vậy giá trị nhỏ nhất của biểu thức \(P=\frac{1}{3}\)

     \(\)