Lời giải:

Áp dụng BĐT Bunhiacopxky:

$C^2\leq (a+b)[(29a+3b)+(29b+3a)]=32(a+b)^2$

$(a+b)^2\leq (a^2+b^2)(1+1)\leq 4$

$\Rightarrow C^2\leq 32.4$

$\Rightarrow C\leq 8\sqrt{2}$
Vậy $C_{\max}=8\sqrt{2}$. Dấu "=" xảy ra khi $a=b=1$

\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Lại có:

\(a^2+b^2+c^2\le1\Rightarrow0\le a;b;c\le1\)

\(\Leftrightarrow a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)\le0\)

\(\Leftrightarrow a+b+c\ge a^2+b^2+c^2=1\)

Do đó:

\(Q^2=2\left(a+b+c\right)+2\sqrt{a^2+ab+bc+ca}+2\sqrt{b^2+ab+bc+ca}+2\sqrt{c^2+ab+bc+ca}\)

\(Q^2\ge2\left(a+b+c\right)+2\sqrt{a^2}+2\sqrt{b^2}+2\sqrt{c^2}\)

\(Q^2\ge4\left(a+b+c\right)\ge4\)

\(\Rightarrow Q\ge2\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị

Áp dụng BĐT Bunhiacopxki ta có: 

\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)

\(=3\left(2a+2b+2c\right)=3.2\left(a+b+c\right)=6.2021=12126\)

\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{12126}\)

Dấu ''='' xảy ra khi \(a=b=c=\dfrac{2021}{3}\)

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

Aps dụng Bunhia-cốpxki : \(P^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\le\left(1+1+1\right)\left(a+b+b+c+c+a\right)\)

\(=6\left(a+b+c\right)\)

\(=6.2021=12126\Leftrightarrow P=\sqrt{12126}\)

Vậy \(Max\left(P\right)=\sqrt{12126}\Leftrightarrow a=b=c=\dfrac{2021}{3}\)

(Refer ;-;)

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

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 \(Min\left(P\right)=2\sqrt{2021}\Leftrightarrow\hept{\begin{cases}a=2021\\b=c=0\end{cases}}\)

\(9=3a^2+2b^2+2bc+2c^2=\left(a+b+c\right)^2+2a^2+b^2+c^2-2a\left(b+c\right)\)

\(\Rightarrow9\ge\left(a+b+c\right)^2+2a^2+\dfrac{1}{2}\left(b+c\right)^2-2a\left(b+c\right)\)

\(\Rightarrow9\ge\left(a+b+c\right)^2+\dfrac{1}{2}\left(2a-b-c\right)^2\ge\left(a+b+c\right)^2\)

\(\Rightarrow-3\le a+b+c\le3\)

\(T_{max}=3\) khi \(a=b=c=1\)

\(T_{min}=-3\) khi \(a=b=c=-1\)