Ta có kết quả tổng quát hơn như sau:

Cho $a,b,c \neq 0$ thỏa mãn $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0.$

Chứng minh rằng $$S={\frac {k{a}^{2}-k-1}{{a}^{2}+2\,bc}}+{\frac {{b}^{2}k-k-1}{2\,ac+{b}^{2}}}+{\frac {{c}^{2}k-k-1}{2\,ab+{c}^{2}}}=k$$

Ta có : 

\(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\frac{3^2}{\left(a+b+c\right)^2}\)

\(A\ge\frac{9}{3^2}=1\)Dấu "=" khi  a=b=c= 1

Bài 4:

Áp dụng bất đẳng thức Cauchy-shwarz dạng engel ta có:

\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}=\dfrac{9}{9}=1\)

Dấu " = " xảy ra khi a = b = c = 1

\(\Rightarrowđpcm\)

Bài 1:
Ta có:
\(a^2+b^2-\frac{(a+b)^2}{2}=\frac{2(a^2+b^2)-(a+b)^2}{2}=\frac{(a-b)^2}{2}\geq 0\)

\(\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}=\frac{2^2}{2}=2\)

(đpcm)

Dấu "=" xảy ra khi $a=b=1$

Bài 2:

\(P-(ab+bc+ac)=\frac{2a^2+2b^2+2c^2-2ab-2bc-2ac}{2}\)

\(=\frac{(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)}{2}=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0, \forall a,b,c\)

\(\Rightarrow P\geq ab+bc+ac\Leftrightarrow P\geq 12\)

Vậy GTNN của $P$ là $12$ khi $a=b=c=2$

\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)

\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)