2. Áp dụng bất đẳng thức Cô - si cho 3 số dương \(\frac{a}{b},\frac{b}{c},\frac{c}{a}\)ta có

\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}\)\(=3\)

Dấu "=" xảy ra <=> a = b = c

xin chào nha mk ko bt đâu ok

Áp dụng BĐT cô si với hai số không âm ta có :

\(1.\sqrt{a+1}\le\frac{a+1+1}{2}=\frac{a}{2}+1\)

\(1.\sqrt{b+1}\le\frac{b}{2}+1\)

\(1.\sqrt{c+1}\le\frac{c}{2}+1\)

=> \(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le3+\frac{a+b+c}{2}=3+\frac{1}{2}=3,5\)

=> ĐPCM 

\(A=\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\Rightarrow A^2=\left(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\right)^2\)

\(\Rightarrow A^2\le\left(1+1+1\right)\left(\sqrt{a+1}^2+\sqrt{b+1}^2+\sqrt{c+1}^2\right)\left(bunhiacopxki\right)\)

\(\Rightarrow A^2\le\left(1+1+1\right)\left(a+1+b+1+c+1\right)\)

\(\Rightarrow A^2\le3\left(a+b+c+3\right)=3.4=12\Rightarrow A\le\sqrt{12}< 3,5\left(dpcm\right)\)

\(a.\) Áp dụng BĐT Cô - Si cho các số không âm , ta có :

\(\sqrt{1}.\sqrt{a+1}\le\dfrac{a+1+1}{2}=\dfrac{a+2}{2}\)

\(\sqrt{1}.\sqrt{b+1}\le\dfrac{b+1+1}{2}=\dfrac{b+2}{2}\)

\(\sqrt{1}.\sqrt{c+1}\le\dfrac{c+1+1}{2}=\dfrac{c+2}{2}\)

\(\Rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le\dfrac{a+b+c+6}{2}=\dfrac{7}{2}=3,5\)

Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}a+1=1\\b+1=1\\c+1=1\end{matrix}\right.\)\(\Leftrightarrow a=b=c=0\)\(\Rightarrow a+b+c\ne1\left(trái-với-giả-thiết\right)\)

\(\Rightarrow\) Dấu \("="\) không xảy ra .

\(\Rightarrow\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}< 3,5\)

\(b.\) Áp dụng BĐT Bunhiacopxki , ta có :

\(\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+a+c\right)=3.2=6\)

\(\Rightarrow\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}\le\sqrt{6}\)

Dấu " = " xảy ra khi : \(a+b=b+c=a+c\Rightarrow a=b=c=\dfrac{1}{3}\)

Câu a : Dùng BĐT Bu-nhi-a-cốp-xki ta có :

\(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le\sqrt{3\left(a+b+c+3\right)}=\sqrt{12}=3,46< 3,5\)

Câu b tương tự :

\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6\left(a+b+c\right)}=\sqrt{6}\)