Lời giải:

Do $a\geq 4, b\geq 5, c\geq 6$

$\Rightarrow c^2=90-a^2-b^2\leq 90-4^2-5^2=49$

$\Rightarrow c\leq 7$

$a^2=90-b^2-c^2\leq 90-5^2-6^2=29< 81$

$\Rightarrow a< 9$

$b^2=90-a^2-c^2=90-4^2-6^2=38< 64$

$\Rightarrow b< 8$

Vậy $4\leq a< 9, 5\leq b< 8, 6\leq c\leq 7$

Suy ra:

$(a-4)(a-9)\leq 0$

$(b-5)(b-8)\leq 0$

$(c-6)(c-7)\leq 0$

$\Rightarrow (a-4)(a-9)+(b-5)(b-8)+(c-6)(c-7)\leq 0$

$\Rightarrow a^2+b^2+c^2+118\leq 13(a+b+c)$

$\Rightarrow 90+208\leq 13P$
$\Rightarrow P\geq 16$

Vậy $P_{\min}=16$. Giá trị này đạt tại $(a,b,c)=(4,5,7)$

fkfkbang14

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

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

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

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

\(T_{max}=3\) khi \(a=b=c=1\)

\(T_{min}=-3\) khi \(a=b=c=-1\)

\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow-3\le a+b+c\le3\)

\(S=a+b+c+\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\dfrac{1}{2}\left(a+b+c\right)^2+a+b+c-\dfrac{3}{2}\)

Đặt \(a+b+c=x\Rightarrow-3\le x\le3\)

\(S=\dfrac{1}{2}x^2+x-\dfrac{3}{2}=\dfrac{1}{2}\left(x+1\right)^2-2\ge-2\)

\(S_{min}=-2\) khi \(\left\{{}\begin{matrix}a+b+c=-1\\a^2+b^2+c^2=3\end{matrix}\right.\) (có vô số bộ a;b;c thỏa mãn)

\(S=\dfrac{1}{2}\left(x^2+2x-15\right)+6=\dfrac{1}{2}\left(x-3\right)\left(x+5\right)+6\le6\)

\(S_{max}=6\) khi \(x=3\) hay \(a=b=c=1\)

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

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