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\(P=\frac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{3\left(a+b\right)}{4\sqrt{ab}}+\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
\(\Rightarrow P\ge\frac{3.2\sqrt{ab}}{4\sqrt{ab}}+2\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}=\frac{3}{2}+1=\frac{5}{2}\)
\(\Rightarrow P_{min}=\frac{5}{2}\) 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
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: \(P=\frac{a^2+3ab+b^2}{\sqrt{ab}\left(a+b\right)}=\frac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
\(=\frac{3\left(a+b\right)}{4\sqrt{ab}}+\left(\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\right)\ge\frac{3.2\sqrt{ab}}{4\sqrt{ab}}+2=\frac{6}{4}+2=2,5\)Vậy Min P = 2,5 đạt được khi a = b
Tại sao \(\frac{3\left(a+b\right)}{4\sqrt{ab}}+\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\) ≥ \(\frac{3.2\sqrt{ab}}{4\sqrt{ab}}\) + 2 vậy ???
\(P=\dfrac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}=\dfrac{a+b}{\sqrt{ab}}+\dfrac{\sqrt{ab}}{a+b}=\dfrac{3\left(a+b\right)}{4\sqrt{ab}}+\dfrac{a+b}{4\sqrt{ab}}+\dfrac{\sqrt{ab}}{a+b}\)
\(\Rightarrow P\ge\dfrac{3.2\sqrt{ab}}{4\sqrt{ab}}+2\sqrt{\dfrac{a+b}{4\sqrt{ab}}.\dfrac{\sqrt{ab}}{a+b}}=\dfrac{3}{2}+1=\dfrac{5}{2}\)
\(\Rightarrow P_{min}=\dfrac{5}{2}\) khi \(a=b\)
thanks