\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
áp dụng bunhia - cốpxki
\(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)
\(=6\left(a+b+c\right)\)
\(=6.2021=12126< =>P=\sqrt{12126}\)
vậy MAX P=\(\sqrt{12126}\)
\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
\(\Rightarrow P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
Áp dụng BĐT Bunyakovsky ta có:
\(P^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\left(a+b+c\right)=6\cdot2021\)
\(\Rightarrow P\le\sqrt{6\cdot2021}=\sqrt{12126}\)
Dấu "=" xảy ra khi: \(a=b=c=\frac{2021}{3}\)
Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\frac{2021}{3}\)
Ta có: \(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)
\(=2\left(a+b+c\right)+2\left[\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(c+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\right]\)
\(=4042+2\left[\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(b+c\right)\left(c+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\right]\)
Mà \(\left(a+b\right)\left(b+c\right)\ge\left(0+b\right)\left(b+0\right)=b^2\)
và \(\left(b+c\right)\left(c+a\right)\ge c^2\) ; \(\left(c+a\right)\left(a+b\right)\ge a^2\)
\(\Rightarrow P\ge4042+2\left(a+b+c\right)=4042+4042=8084\)
\(\Rightarrow P\ge2\sqrt{2021}\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}a=2021\\b=c=0\end{cases}}\) và các hoán vị của nó
Vậy \(Min\left(P\right)=2\sqrt{2021}\Leftrightarrow\hept{\begin{cases}a=2021\\b=c=0\end{cases}}\)
