Ta có: \(2\left(b^2+bc+c^2\right)=2b^2+2c^2+2bc\le2b^2+2c^2+b^2+c^2=3\left(b^2+c^2\right)\Rightarrow b^2+c^2\le3-a^2\Rightarrow a^2+b^2+c^2\le3\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\).

Áp dụng bđt Schwars ta có:

\(T\ge a+b+c+\dfrac{18}{a+b+c}=\left(a+b+c+\dfrac{9}{a+b+c}\right)+\dfrac{9}{a+b+c}\ge2\sqrt{9}+\dfrac{9}{3}=9\).

Đẳng thức xảy ra khi a = b = c = 1.

Áp dụng bất đẳng thức Cauchy - Schwarz dưới dạng Engel ta có :

\(A=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{1+1+1}=\frac{2^2}{3}=\frac{4}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{2}{3}\)

Vậy .............

Ta dễ có BĐT sau \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Khi đó \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{4}{3}\)

Đẳng thức xảy ra tại a=b=c=²/₃

\(\dfrac{4}{3}=a+2\sqrt{\dfrac{a}{4}.b}+\dfrac{1}{2}\sqrt[3]{\dfrac{a}{2}.2b.8c}\)

\(\dfrac{4}{3}\le a+\dfrac{a}{4}+b+\dfrac{1}{6}\left(\dfrac{a}{2}+2b+8c\right)=\dfrac{4}{3}\left(a+b+c\right)\)

\(\Rightarrow a+b+c\ge1\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)

Ta có: \(a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)=0\)

\(\Leftrightarrow\)\(a\left(a-b\right)-b\left(a-b+c-a\right)+c\left(c-a\right)=0\)

\(\Leftrightarrow\)\(a\left(a-b\right)-b\left(a-b\right)-b\left(c-a\right)+c\left(c-a\right)=0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(c-a\right)\left(c-b\right)=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\\left(c-a\right)\left(c-b\right)=0\end{cases}}\)

\(\Leftrightarrow\)\(a=b=c\)

Thế a = b = c vào A ta được:

\(A=3^3-3a^3+3a^2-3a+5\)

\(A=3\left(a^2-a+\frac{5}{3}\right)\)

\(A=3\left[\left(a-\frac{1}{2}\right)^2+\frac{17}{12}\right]\)

\(A=3\left(a-\frac{1}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)

Vậy GTNN của A là 17/4 khi a = b = c = 1/2

Ta có: \(a\left(a-b\right)+b\left(b-c\right)+c\left(c-a\right)=0\)

<=> \(a^2+b^2+c^2-ac-bc-ab=0\Leftrightarrow2a^2+2b^2+2c^2-2ac-2bc-2ab=0\)

<=> \(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

<=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

<=> \(\left(a-b\right)^2=0,\left(b-c\right)^2=0,\left(a-c\right)^2=0\)

<=> a=b=c

Thế vào ta có biểu thức:

A=\(3a^3-3a^3+3a^2-3a+5=3\left(a^2-a+\frac{5}{3}\right)=3\left(a-\frac{1}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\)

Giá trị nhỏ nhất của biểu thức A=17/4 

Dấu bằng xảy ra khi a=b=c=1/2

\(A=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\\ A=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\right)\\ A=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\dfrac{1}{4}\left(a+2b+3c\right)\\ A\ge2\sqrt{\dfrac{3a}{4}\cdot\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}\cdot\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}\cdot\dfrac{4}{c}}+\dfrac{1}{4}\cdot20\\ A\ge3+3+2+5=13\\ A_{min}=13\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)

làm lại dong cuối:\(A\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{a}\)

Mà:\(2c+b=abc\Rightarrow a=\frac{2c+b}{cb}=\frac{2}{b}+\frac{1}{c}\)

\(\Rightarrow2a=\frac{4}{b}+\frac{2}{c}\)

\(\Rightarrow A\ge2a+\frac{6}{a}\)

Ta có:\(A=\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}\right)+2\left(\frac{1}{b+c-a}+\frac{1}{a+b-c}\right)\)

\(+3\left(\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\)

\(\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{c}\) (Do a,b,c là 3 cạnh của tam giác nên:\(\hept{\begin{cases}a+b-c>0\\a+c-b>0\\c+b-a>0\end{cases}}\)

\(=\frac{6}{a}+2a\ge4\sqrt{3}\left(cosi\right)\left(a>0\right)\)

Dấu = xảy ra khi:

\(a=b=c=\sqrt{3}\)

xin lỗi các bạn đáp án là\(2\sqrt{3}\)

đề sai

Ta có: \(\left(a+b\right)^2\ge4ab=16\Rightarrow a+b\ge4\Rightarrow a+b-4\ge0\)

\(P=\dfrac{1+b+1+a}{\left(1+a\right)\left(1+b\right)}=\dfrac{a+b+2}{ab+a+b+1}=\dfrac{a+b+2}{a+b+5}\)

\(P=\dfrac{3a+3b+6}{3\left(a+b+5\right)}=\dfrac{2\left(a+b+5\right)+\left(a+b-4\right)}{3\left(a+b+5\right)}\ge\dfrac{2\left(a+b+5\right)}{3\left(a+b+5\right)}=\dfrac{2}{3}\)

\(P_{min}=\dfrac{2}{3}\) khi \(a=b=2\)