\(a+b+c=3\\ \Leftrightarrow a\left(b+c+2\right)=ab+ac+a+b+c+1=\left(a+1\right)\left(b+c+1\right)\)

Tương tự:

\(b\left(c+a+2\right)=\left(b+1\right)\left(a+c+1\right)\\ c\left(a+b+2\right)=\left(c+1\right)\left(a+b+1\right)\)

Áp dụng BĐT cosi:

\(\left\{{}\begin{matrix}\left(a+1\right)\left(b+c+1\right)\le\dfrac{\left(a+1+b+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(b+1\right)\left(a+c+1\right)\le\dfrac{\left(b+1+a+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(c+1\right)\left(a+b+1\right)\le\dfrac{\left(c+1+a+b+1\right)^2}{2}=\dfrac{2^2}{2}=2\end{matrix}\right.\)

Cộng vế theo vế 2 BĐT trên:

\(\Leftrightarrow\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(c+a+2\right)}+\sqrt{c\left(a+b+2\right)}\le2+2+2=6\)

Dấu \("="\Leftrightarrow a=b=c=1\)

Áp dụng BĐT Bunhiacopski:

\(VT^2=\left(\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(a+c+2\right)}+\sqrt{c\left(a+b+2\right)}\right)^2\\ \le\left(a+b+c\right)\left(b+c+2+a+c+2+a+b+2\right)\\ =3\cdot\left(2\cdot3+6\right)=36\\ \Leftrightarrow VT\le\sqrt{36}=6\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{b+c+2}}{\sqrt{a}}=\dfrac{\sqrt{a+c+2}}{\sqrt{b}}=\dfrac{\sqrt{a+b+2}}{\sqrt{c}}\\a+b+c=3\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=1\)

a)Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\)

\(\le2\cdot\left(1+1+1\right)\left(a+b+c\right)\le6\)

\(\Rightarrow VT^2\le6\Rightarrow VT\le\sqrt{6}=VP\)

Xảy ra khi \(a=b=c=\frac{1}{3}\)

b)Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{a+\sqrt{b+\sqrt{2c}}}+\sqrt{b+\sqrt{c+\sqrt{2a}}}+\sqrt{c+\sqrt{a+\sqrt{2b}}}\right)^2\)

\(\le\left(1+1+1\right)\left(a+b+c+Σ\sqrt{b+\sqrt{2c}}\right)\)

\(=3\left(6+\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)\)

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

\(\le\left(1+1+1\right)\left(a+b+c+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\)

\(=3\left(6+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\)

Đặt tiếp: \(B^2=\left(\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)^2\)

\(\le2\cdot\left(1+1+1\right)\left(a+b+c\right)\le36\Rightarrow B\le6\)

\(\Rightarrow A^2\le3\left(6+\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\right)\le3\cdot12=36\Rightarrow A\le6\)

\(\Rightarrow VT^2\le3\left(6+\sqrt{b+\sqrt{2c}+\sqrt{c+\sqrt{2a}}}+\sqrt{a+\sqrt{2b}}\right)\)

\(\le3\left(6+6\right)=3\cdot12=36\Rightarrow VT\le6=VP\)

Xảy ra khi \(a=b=c=2\)

đặt A=...

Áp dúng bất đẳng thức bu nhi a ta có 

\(A^2\le3\left(1+a^2+2bc+1+b^2+2ac+1+c^2+2ab\right)=3\left[\left(a+b+c\right)^2+3\right]\)

=> \(A^2\le36\Rightarrow A\le6\) (ĐPCM)

dấu = xảy ra <=> a=b=c=1

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

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

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

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

Theo BĐT Bu nhi a cốp xki ta có :

\(VT=\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\le\sqrt{3\left(4a+4b+4c\right)}=\sqrt{12\left(a+b+c\right)}=\sqrt{36}=6\)

Vậy đpcm . Dấu bằng xảy ra khi \(a=b=c=1\)