Question

Cho a,b,c >0 t/m a+b+c=abc-2. Tìm max

\(P=\sqrt{\dfrac{1}{a+1}}+\sqrt{\dfrac{1}{b+1}}+\sqrt{\dfrac{1}{c+1}}\)

Answer 1

\(a+b+c+2=abc\)

\(\Leftrightarrow2a+2b+2c+3+ab+bc+ca=abc+ab+bc+ca+a+b+c+1\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)+\left(c+1\right)\left(b+1\right)+\left(c+1\right)\left(a+1\right)=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)

\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=1\)

Đặt \(\left(\dfrac{1}{a+1};\dfrac{1}{b+1};\dfrac{1}{c+1}\right)=\left(x;y;z\right)\)

\(\Rightarrow x+y+z=1\)

BĐT trở thành:

\(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{3}\) hay \(a=b=c=2\)