Ta có :

\(p=\sqrt{\dfrac{ab}{\left(1-a\right)\left(1-b\right)}}+\sqrt{\dfrac{bc}{\left(1-b\right)\left(1-c\right)}}+\sqrt{\dfrac{ac}{\left(1-a\right)\left(1-c\right)}}\)

Theo Bất Đẳng Thức Cosi ta có :

\(2\sqrt{\dfrac{ab}{\left(1-a\right)\left(1-b\right)}}=2\sqrt{\dfrac{a}{1-b}\times\dfrac{b}{1-a}}\le\dfrac{a}{1-b}+\dfrac{b}{1-a}\)

\(2\sqrt{\dfrac{bc}{\left(1-b\right)\left(1-c\right)}}=2\sqrt{\dfrac{b}{1-c}\times\dfrac{c}{1-b}}\le\dfrac{b}{1-c}+\dfrac{c}{1-b}\)

\(2\sqrt{\dfrac{ac}{\left(1-a\right)\left(1-c\right)}}=2\sqrt{\dfrac{a}{1-c}\times\dfrac{c}{1-a}}\le\dfrac{a}{1-c}+\dfrac{c}{1-a}\)

Cộng vế với vế ta được :

\(2p\le\dfrac{a+c}{1-b}+\dfrac{b+c}{1-a}+\dfrac{a+b}{a-c}\)

\(\Leftrightarrow2p\le\dfrac{1-b}{1-b}+\dfrac{1-a}{1-a}+\dfrac{1-c}{1-c}\)

\(\Leftrightarrow2p\le3\)

\(\Leftrightarrow p\le\dfrac{3}{2}\)

Mặt khác a = b = c = \(\dfrac{1}{3}\) thì \(p=\dfrac{3}{2}\)

Vậy maxP = \(\dfrac{3}{2}\)

p = √one b( 1 - a ) ( 1 - b ) +√b c( 1 - b ) ( 1 - c ) +√một c( 1 - a ) ( 1 - c )

Theo Bất Đẳng Thức Cosi ta có:

2 √one b( 1 - a ) ( 1 - b ) =2√một1 - b ×b1 - a ≤a1 - b +b1 - a

2 √b c( 1 - b ) ( 1 - c ) =2√b1 - c ×c1 - b ≤b1 - c +c1 - b

2 √một c( 1 - a ) ( 1 - c ) =2√một1 - c ×c1 - a ≤a1 - c +c1 - a

Cộng sản:

2 p ≤ a + c1 - b +b+c1 - a +a+ba - c

⇔ 2 p ≤ 1 - b1 - b +1-a1 - a +1-c1-c

⇔2p≤3

⇔p≤32

Mặt other a = b = c =  13  thì p=32

Then maxP =  32