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\(\sqrt{2}A=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(a+1\right)}\le\frac{2a+2b+a+b+2}{2}=\frac{8}{2}=4\)
\(\Rightarrow A\le\frac{4}{\sqrt{2}}=2\sqrt{2}.\text{Dấu "=" xảy ra khi:}a=b=1\)
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 \(\frac{4\sqrt{2}}{3}+\frac{1}{2}\Leftrightarrow a=b=\frac{1}{3}\)
Đặt: \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(S=\frac{\frac{1}{x}}{\sqrt{\frac{1}{y}.\frac{1}{z}\left(1+\frac{1}{x^2}\right)}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{z}.\frac{1}{x}\left(1+\frac{1}{y^2}\right)}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{x}.\frac{1}{y}\left(1+\frac{1}{z^2}\right)}}\)
\(=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
\(=\sqrt{\frac{yz}{xy+yz+zx+x^2}}+\sqrt{\frac{zx}{xy+yz+zx+y^2}}+\sqrt{\frac{xy}{xy+yz+zx+z^2}}\)
\(=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{zx}{\left(y+x\right)\left(y+z\right)}}+\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\)
\(\le\frac{1}{2}.\left(\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{x}{z+x}+\frac{y}{z+y}\right)\)
\(=\frac{1}{2}.\left(1+1+1\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\)
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
\(P=\frac{\left(a+b\right)^2+ab}{\sqrt{ab}\left(a+b\right)}=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{3\left(a+b\right)}{4\sqrt{ab}}+\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
\(\Rightarrow P\ge\frac{3.2\sqrt{ab}}{4\sqrt{ab}}+2\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}=\frac{3}{2}+1=\frac{5}{2}\)
\(\Rightarrow P_{min}=\frac{5}{2}\) khi a=b
Vì ( a - b )2 \(\ge\)0 \(\forall\)a,b \(\Rightarrow a^2+b^2\ge2ab\). Mà ab = 4 \(\Rightarrow a^2+b^2\ge8\)
\(\Rightarrow\frac{\left(a+b-2\right)\left(a^2+b^2\right)}{a+b}\ge\frac{\left(a+b-2\right).8}{a-b}\)
Đặt t = a + b \(\Rightarrow t\ge4\)( Do \(a+b\ge2\sqrt{ab}=4\))
\(\frac{\left(t-2\right).8}{t}=\frac{8t-16}{t}=8-\frac{16}{t}\)
Vì \(t\ge4\Rightarrow\frac{16}{t}\le\frac{16}{4}\Rightarrow-\frac{16}{t}\ge-4\Rightarrow\left(8-\frac{16}{t}\right)\ge8-4=4\)
\(\Rightarrow\frac{\left(a+b-2\right)\left(a^2+b^2\right)}{a+b}\ge4\)Dấu '' = '' xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\a,b=4\end{cases}\Leftrightarrow a=b=2}\)
Vậy \(\frac{\left(a+b-2\right)\left(a^2+b^2\right)}{a+b}\)min \(\Leftrightarrow a=b=2\)
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 ???