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Ta có: \(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)
\(\Rightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}\Leftrightarrow\orbr{\begin{cases}a^2+2ab+b^2\ge4ab\\2\left(a^2+b^2\right)\ge a^2+2ab+b^2\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(a+b\right)^2\ge4ab\left(1\right)\\\left(a+b\right)^2\le2\left(a^2+b^2\right)\left(2\right)\end{cases}}\)
Theo đề bài:
\(a+b+3ab=1\)
\(\Leftrightarrow4\left(a+b\right)+12ab=4\)
\(\Leftrightarrow4\left(a+b\right)+3\left(a+b\right)^2\ge4\left(theo\left(1\right)\right)\)
\(\Leftrightarrow3\left(a+b\right)^2+4\left(a+b\right)-4\ge0\)
\(\Leftrightarrow\left(a+b+2\right)\left[3\left(a+b\right)-2\right]\ge0\)
\(\Leftrightarrow3\left(a+b\right)-2\ge0\left(a,b>0\Rightarrow a+b+2>0\right)\)
\(\Leftrightarrow a+b\ge\frac{2}{3}\)
`\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\ge\frac{4}{9}\left(theo\left(2\right)\right)\)
Áp dụng các kết quả trên, ta có:
\(\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2\le2\left(1-a^2+1-b^2\right)\)\(=4-2\left(a^2+b^2\right)\le4-\frac{4}{9}=\frac{32}{9}\)
\(\Rightarrow\sqrt{1-a^2}+\sqrt{1-b^2}\le\frac{4\sqrt{2}}{3}\)
Ta có: \(\frac{3ab}{a+b}=\frac{1-\left(a+b\right)}{a+b}=\frac{1}{a+b}-1\le\frac{1}{\frac{2}{3}}-1=\frac{1}{2}\)
\(\Rightarrow A\le\frac{4\sqrt{2}}{3}+\frac{1}{2}\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b+3ab=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\3a^2+2a-1=0\end{cases}\Leftrightarrow}a=b=\frac{1}{3}\left(a,b>0\right)}\)
Vậy max A là \(\frac{4\sqrt{2}}{3}+\frac{1}{2}\Leftrightarrow a=b=\frac{1}{3}\)
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 ???
Áp dụng bđt AM-GM ta có
\(P\ge\frac{4}{2+a^2+b^2+6ab}=\frac{4}{\left(a+b\right)^2+4ab+1}=\frac{2}{1+2ab}\)
Lại có \(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)
\(\Rightarrow P\ge\frac{2}{1+\frac{1}{2}}=\frac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
cảm ơn
Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có :
\(A=\frac{1}{1+3ab+a^2}+\frac{1}{1+3ab+b^2}\ge\frac{4}{a^2+b^2+6ab+2}\)
Ta có : \(a^2+b^2+6ab+2=\left(a^2+2ab+b^2\right)+4ab+2=\left(a+b\right)^2+4ab+2=4ab+3\)
Áp dụng bđt \(xy\le\frac{\left(x+y\right)^2}{4}\) ta có : \(4ab+3\le4.\frac{\left(a+b\right)^2}{4}+3=\left(a+b\right)^2+3=1+3=4\)
\(\Rightarrow A\ge\frac{4}{a^2+b^2+6ab+2}\ge\frac{4}{4}=1\) có GTNN là 1
Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)