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\(A=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{\left(ab\right)^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{\left(ab\right)^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{ab}\)
\(=\frac{1}{\left(a+b\right)^3}.\frac{a^3+b^3}{1^3}+\frac{3}{\left(a+b\right)^4}.\frac{a^2+b^2}{1^2}+\frac{6}{\left(a+b\right)^5}.\frac{a+b}{1}\)
\(=\frac{a^2-ab+b^2}{\left(a+b\right)^2}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6}{\left(a+b\right)^4}\)\(=\frac{\left(a^3+b^3\right)\left(a+b\right)+3a^2+3b^2+6}{\left(a+b\right)^4}\)
\(=\frac{a^4+a^3b+ab^3+b^4+3a^2+3b^2+6}{a^4+4a^3b+6a^2b^2+4ab^3+b^4}\)\(=\frac{a^4+a^2.1+1.b^2+b^4+3a^2+3b^2+6}{a^4+4a^2.1+6.1^2+4b^2.1+b^4}\)
\(=\frac{a^4+4a^2+4b^2+b^4+6}{a^4+4a^2+6+4b^2+b^4}=1\)
Ta có : \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{a^3b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{a^2b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{ab}\right)\)
=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{1}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{1}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{1}\right)\)
=> \(P=\frac{a^3+b^3}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2+2a\right)-6a}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a+b\right)^2}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}-\frac{6}{\left(a+b\right)^4}\)
=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}+\frac{6}{\left(a+b\right)^4}-\frac{6}{\left(a+b\right)^4}\)
=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}=\frac{2a^2+4ab+2b^2}{\left(a+b\right)^2}-\frac{a^2+b^2}{\left(a+b\right)^2}\)
=> \(P=2-\frac{a^2+b^2}{\left(a+b\right)^2}=1+\frac{-2}{\left(a+b\right)^2}\)
Đặt \(\frac{1}{a}=x>0;\frac{1}{b}=y>0;\frac{1}{c}=z>0\)
Từ giả thiết ta có: \(7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\le6\left(x^2+y^2+z^2\right)+2015\)
\(\Leftrightarrow x^2+y^2+z^2\le2015\)
Ta có: \(\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}=\frac{1}{\sqrt{\left(4a^2+b^2\right)+\left(2a^2+2b^2\right)}}\le\frac{1}{\sqrt{4a^2+b^2+4ab}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(2x+y\right)\)
Tương tự thì: \(\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\frac{1}{9}\left(2y+z\right)\) và \(\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{1}{9}\left(2z+x\right)\)
Cộng từng vế 3 BĐT trên ta có: \(\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{x+y+z}{3}\le\frac{\sqrt{3\left(x^2+y^2+z^2\right)}}{3}\le\sqrt{\frac{2015}{3}}\)
Vậy max \(P=\sqrt{\frac{2015}{3}}\) , đạt được khi \(a=b=c=\sqrt{\frac{3}{2015}}\)
Bài 1:Với \(ab=1;a+b\ne0\) ta có:
\(P=\frac{a^3+b^3}{\left(a+b\right)^3\left(ab\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4\left(ab\right)^2}+\frac{6\left(a+b\right)}{\left(a+b\right)^5\left(ab\right)}\)
\(=\frac{a^3+b^3}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)
\(=\frac{a^2+b^2-1}{\left(a+b\right)^2}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6}{\left(a+b\right)^4}\)
\(=\frac{\left(a^2+b^2-1\right)\left(a+b\right)^2+3\left(a^2+b^2\right)+6}{\left(a+b\right)^4}\)
\(=\frac{\left(a^2+b^2-1\right)\left(a^2+b^2+2\right)+3\left(a^2+b^2\right)+6}{\left(a+b\right)^4}\)
\(=\frac{\left(a^2+b^2\right)^2+4\left(a^2+b^2\right)+4}{\left(a+b\right)^4}=\frac{\left(a^2+b^2+2\right)^2}{\left(a+b\right)^4}\)
\(=\frac{\left(a^2+b^2+2ab\right)^2}{\left(a+b\right)^4}=\frac{\left[\left(a+b\right)^2\right]^2}{\left(a+b\right)^4}=1\)
Bài 2: \(2x^2+x+3=3x\sqrt{x+3}\)
Đk:\(x\ge-3\)
\(pt\Leftrightarrow2x^2-3x\sqrt{x+3}+\sqrt{\left(x+3\right)^2}=0\)
\(\Leftrightarrow2x^2-2x\sqrt{x+3}-x\sqrt{x+3}+\sqrt{\left(x+3\right)^2}=0\)
\(\Leftrightarrow2x\left(x-\sqrt{x+3}\right)-\sqrt{x+3}\left(x-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+3}\right)\left(2x-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=x\\\sqrt{x+3}=2x\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\ge0\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-x-3=0\left(x\ge0\right)\\4x^2-x-3=0\left(x\ge0\right)\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{13}}{2}\\x=1\end{cases}\left(x\ge0\right)}\)
Bài 4:
Áp dụng BĐT AM-GM ta có:
\(2\sqrt{ab}\le a+b\le1\Rightarrow b\le\frac{1}{4a}\)
Ta có: \(a^2-\frac{3}{4a}-\frac{a}{b}\le a^2-\frac{3}{4a}-4a^2=-\left(3a^2+\frac{3}{4a}\right)\)
\(=-\left(3a^2+\frac{3}{8a}+\frac{3}{8a}\right)\le-3\sqrt[3]{3a^2\cdot\frac{3}{8a}\cdot\frac{3}{8a}}=-\frac{9}{4}\)
Đẳng thức xảy ra khi \(a=b=\frac{1}{2}\)