Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Đây là bài sử dụng Cô-si ngược dấu đặc trưng:

\(\dfrac{1}{a^2+1}=\dfrac{a^2+1-a^2}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)

Tương tự: \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2}\); 

\(\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)

Cộng vế:

\(P\ge3-\dfrac{a+b+c}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz và AM-GM:

$M=\frac{b^2+c^2}{a^2}+a^2(\frac{1}{b^2}+\frac{1}{c^2})$

$\geq \frac{b^2+c^2}{a^2}+a^2.\frac{4}{b^2+c^2}$

$=(\frac{b^2+c^2}{a^2}+\frac{a^2}{b^2+c^2})+\frac{3a^2}{b^2+c^2}$

$\geq \sqrt{\frac{b^2+c^2}{a^2}.\frac{a^2}{b^2+c^2}}+\frac{3(b^2+c^2)}{b^2+c^2}$

$=2+3=5$

Vậy $M_{\min}=5$ 

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)

                             LG

Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)

                                  \(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)

                                  \(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)

                                 \(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)

                                 \(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)

Khi đó :\(B=a+b+c+\frac{1}{abc}\)

   \(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)

\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)

 \(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)

Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)

Vậy .........

2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)

\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)

\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)

Áp dụng BĐT AM-GM ta có:

\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)

\(A\ge a+b+c-\frac{6}{2}\)

\(A\ge6-3\)

\(A\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)

                                 \(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)

                                 \(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)

Lấy \(\left(1\right)-\left(3\right)\)ta có:

\(2a-2c=c+b-a-b=c-a\)

\(\Rightarrow2a-2c-c+a=0\)

\(\Leftrightarrow3.\left(a-c\right)=0\)

\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)

Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)

\(\Rightarrow a=b=c=2\)

Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)

Ta có: \(a+b+c+ab+bc+ca=6\)

Áp dụng BĐT AM-GM ta có:

\(a+b+c+ab+bc+ca\ge6.\sqrt[6]{\left(abc\right)^3}\)

\(\Leftrightarrow6\ge6.\sqrt[6]{\left(abc\right)^3}\)

\(\Leftrightarrow1\ge\sqrt[6]{\left(abc\right)^3}\)

\(\Leftrightarrow1\ge\left(abc\right)^3\)

\(\Leftrightarrow1\ge abc\)

\(\Leftrightarrow1\ge C\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=ab=bc=ca\Leftrightarrow a=b=c=1\)

Vậy \(C_{max}=1\Leftrightarrow a=b=c=1\)