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Ta có:
\(P=\sqrt{a\left(b+1\right)}+\sqrt{b\left(a+1\right)}=\sqrt{a}.\sqrt{b+1}+\sqrt{b}.\sqrt{a+1}\)
Áp dụng bđt \(B.C.S\) lần lượt cho hai bộ số thực gồm \(\left(\sqrt{a};\sqrt{b}\right)\) và \(\left(\sqrt{b+1};\sqrt{a+1}\right)\) , ta được:
\(P\le\sqrt{\left(a+b\right)\left[\left(a+1\right)+\left(b+1\right)\right]}\le\sqrt{2\left(2+2\right)}=2\sqrt{2}\) (do \(a+b\le2\) )
Đẳng thức xảy ra khi \(a=b=1\)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
\(P=\sqrt{a\left(b+1\right)}+\sqrt{b\left(a+1\right)}\)
\(\Rightarrow P\sqrt{2}=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(a+1\right)}\)
\(\le\frac{1}{2}\left(2a+b+1\right)+\frac{1}{2}\left(2b+a+1\right)\)
\(\le\frac{1}{2}\left(3a+3b+2\right)\le\frac{1}{2}.\left(3.2+2\right)=4\)
\(\Rightarrow p\le2\sqrt{2}\)
Dấu"=" xảy ra \(\Leftrightarrow a=b=1\)
Vậy Max P \(=2\sqrt{2}\)\(\Leftrightarrow a=b=1\)
Có: \(a+b+c+2\sqrt{abc}=1\Rightarrow\hept{\begin{cases}a+2\sqrt{abc}=1-b-c\\b+2\sqrt{abc}=1-a-c\\c+2\sqrt{abc}=1-a-b\end{cases}}\)
\(A=\sqrt{a\left(1-b\right)\left(1-c\right)}+\sqrt{b\left(1-c\right)\left(1-a\right)}+\sqrt{c\left(1-a\right)\left(1-b\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{a\left(1-b-c+bc\right)}+\sqrt{b\left(1-a-c+ac\right)}+\sqrt{c\left(1-a-b+ab\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{a\left(a+2\sqrt{abc}+bc\right)}+\sqrt{b\left(b+2\sqrt{abc}+ac\right)}+\sqrt{c\left(c+2\sqrt{abc}+ab\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{\left(a^2+2a\sqrt{abc}+abc\right)}+\sqrt{\left(b^2+2b\sqrt{abc}+abc\right)}+\sqrt{\left(c^2+2c\sqrt{abc}+abc\right)}-\sqrt{abc}+2015\)
\(A=\sqrt{\left(a+\sqrt{abc}\right)^2}+\sqrt{\left(b+\sqrt{abc}\right)^2}+\sqrt{\left(c+\sqrt{abc}\right)^2}-\sqrt{abc}+2015\)
\(A=a+\sqrt{abc}+b+\sqrt{abc}+c+\sqrt{abc}-\sqrt{abc}+2015\)
\(A=a+b+c+2\sqrt{abc}+2015\)
\(A=1+2015=2016\)
Vậy:....