cho a,b,c là các số thực lớn hơn 1/4 t/m: 1/a+1+1/b+1+1/c+1=2 . tìm gtnn của bt: P=1/4a+1+1/4b+1+1/4c+1

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{1}{a+1} +\frac{1}{b+1} +\frac{1}{c+1} =2\\
\leftrightarrow \frac{( b+1)( c+1) +( a+1)( c+1) +( a+1)( b+1)}{( a+1)( b+1)( c+1)} =2\\
\leftrightarrow ( b+1)( c+1) +( a+1)( c+1) +( a+1)( b+1) =2( a+1)( b+1)( c+1)\\
\leftrightarrow ab+bc+ac+2abc=1
\end{array}$
1=2abc+ab+bc+ac⩾2abc+$\displaystyle 3\sqrt[3]{a^{2} b^{2} c^{2}}$
Đặt t=$\displaystyle \sqrt[3]{abc}$ ta có $\displaystyle 2t^{3} +3t^{2} -1\leqslant 0\leftrightarrow ( 2t-1)( t+1)^{2} \leqslant 0\leftrightarrow t\leqslant \frac{1}{2} \leftrightarrow abc\leqslant \frac{1}{8}$
$\displaystyle \leftrightarrow 12( 1-2abc) \geqslant ( 32abc-1)^{2} \leftrightarrow 2\sqrt{3( 1-2abc)} \geqslant 32abc-1$ $\displaystyle \leftrightarrow 2\sqrt{3( 1-2abc)} +1\geqslant 32abc\leftrightarrow 2( a+b+c) +1\geqslant 32abc$

$\displaystyle \leftrightarrow ( 4a+1)( 4b+1) +( 4b+1)( 4c+1) +( 4a+1)( 4c+1) \geqslant ( 4a+1)( 4b+1)( 4c+1) \leftrightarrow \frac{1}{4a+1} +\frac{1}{4b+1} +\frac{1}{4c+1} \geqslant 1$
Dấu = xảy ra khi và chỉ khi a=b=c=$\displaystyle \frac{1}{2}$