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\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow-3\le a+b+c\le3\)
\(S=a+b+c+\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\dfrac{1}{2}\left(a+b+c\right)^2+a+b+c-\dfrac{3}{2}\)
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 \(a=b=c=1\)
2:
a: =>a^2+2ab+b^2-2a^2-2b^2<=0
=>-(a^2-2ab+b^2)<=0
=>(a-b)^2>=0(luôn đúng)
b; =>a^2+b^2+c^2+2ab+2ac+2bc-3a^2-3b^2-3c^2<=0
=>-(2a^2+2b^2+2c^2-2ab-2ac-2bc)<=0
=>(a-b)^2+(b-c)^2+(a-c)^2>=0(luôn đúng)
Ta có
\(\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)\left(a^2+b^2+c^2\right)\ge\left(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)^2=\frac{9}{16}\)
\(\Rightarrow a^2+b^2+c^2\ge\frac{3}{4}\)
\(\Rightarrow M=4\left(a^2+b^2+c^2\right)\ge3\)
Đạt được khi: \(a=b=c=\frac{1}{2}\)
Ta có:
\(a^2+b^2\ge2ab\) (1)
\(b^2+c^2\ge2bc\) (2)
\(a^2+c^2\ge2ac\) (3)
Cộng từng vế (1);(2);(3)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2ab+2bc+2ac\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=\frac{9}{4}\)
\(\Leftrightarrow4\left(a^2+b^2+c^2\right)\ge3\)
Dấu "=" xãy ra<=>a=b=c=1/2
vậy MinM=3<=>a=b=c=1/2
\(Q\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\le\sqrt{6.\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt{3}}\)
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