a^3+b^3+ab=(a+b)(a^2+b^2-ab)+ab=a^2+b^2

mà 2(a^2+b^2)>=(a+b)²(vì a^2+b^2>=2ab)

\(\Rightarrow\)a^2+b^2>=1/2

a) Áp dụng bất đẳng thức Bnhiacopxki ta có :

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a.1+b.1+c.1\right)^2\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

b) Ta có : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac+bc\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3ab+3bc+3ac\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)

\(\Rightarrow ab+ac+bc\le\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

\(A=\dfrac{1}{4}.4ab\left(a^2+b^2-2ab\right)\le\dfrac{1}{16}\left(4ab+a^2+b^2-2ab\right)=\dfrac{1}{16}\left(a+b\right)^2=\dfrac{1}{16}\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{2-\sqrt{2}}{4};\dfrac{2+\sqrt{2}}{4}\right);\left(\dfrac{2+\sqrt{2}}{4};\dfrac{2-\sqrt{2}}{4}\right)\)

a.

\(P=\frac{6}{x^2-6x+17}\)

Ta thấy: $x^2-6x+17=(x-3)^2+8\geq 8$ với mọi $x\in\mathbb{R}$

$\Rightarrow P=\frac{6}{x^2-6x+17}\leq \frac{6}{8}=\frac{3}{4}$

Vậy $P_{\max}=\frac{3}{4}$. Giá trị này đạt tại $x-3=0\Leftrightarrow x=3$

b/

Ta có:

$6=a^2+b^2-ab=\frac{1}{2}(a^2+b^2)+\frac{1}{2}(a^2+b^2-2ab)$

$=\frac{1}{2}(a^2+b^2)+\frac{1}{2}(a-b)^2\geq \frac{1}{2}(a^2+b^2)$ với mọi $a,b$

$\Rightarrow 12\geq a^2+b^2$
Vậy $P_{\max}=12$. Giá trị này đạt tại $a=b=\pm \sqrt{6}$

\(M=a^2+ab+b^2-3a-3b+2001\)

\(\Rightarrow2M=2a^2+2ab+2b^2-6a-6b+4002\)

\(=\left[\left(a+b\right)^2-2\left(a+b\right).2+4\right]+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+3996\)

\(=\left(a+b-2\right)^2+\left(a-1\right)^2+\left(b-1\right)^2+3996\ge3996\)

\(\Rightarrow M\ge1998\)

\(minM=1998\Leftrightarrow a=b=1\)