theo cauchy schwars ta có
\(\left\{{}\begin{matrix}2\sqrt{a+b}.1\le a+b+1\\2\sqrt{c+b}.1\le c+b+1\\2\sqrt{a+c}.1\le a+c+1\end{matrix}\right.\)
cộng vế theo vế ta đc \(2\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\le2\left(a+b+c\right)+3=5\)
\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\dfrac{5}{2}\)
vậy \(P_{MAX}=\dfrac{5}{2}\)
Áp dụng bđt Cosi :
\(\sqrt{a+b}\le\dfrac{a+b+1}{2}\)
cmtt : \(\sqrt{b+c}\le\dfrac{b+c+1}{2}\)
\(\sqrt{c+a}\le\dfrac{c+a+1}{2}\)
\(\Rightarrow VT\le\dfrac{a+b+1+b+c+1+c+a+1}{2}=\dfrac{2+3}{2}=\dfrac{5}{2}\)
Dấu bằng xảy ra khi a=b=c
\(P^2=\left(\sqrt{a+b}.1+\sqrt{b+c}.1+\sqrt{c+a}.1\right)\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\Rightarrow P\le\sqrt{6}\)
Dấu "=" xảy ra <=> a = b = c = \(\dfrac{1}{3}\)
