\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)

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

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

Đặt \(\left\{{}\begin{matrix}a-2=x\ge0\\b=y\ge0\end{matrix}\right.\) \(\Rightarrow2y+4=\left(x+2\right)y\Rightarrow xy=4\)

\(P=\dfrac{\sqrt{x^2+2x}}{x+1}+\dfrac{\sqrt{y^2+2y}}{y+1}+\dfrac{1}{x+y+2}\)

\(P=\dfrac{\sqrt{2x\left(x+2\right)}}{\sqrt{2}\left(x+1\right)}+\dfrac{\sqrt{2y\left(y+2\right)}}{\sqrt{2}\left(y+1\right)}+\dfrac{1}{x+1+y+1}\)

\(P\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{3x+2}{x+1}+\dfrac{3y+2}{y+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

\(P\le\dfrac{1}{2\sqrt{2}}\left(3-\dfrac{1}{x+1}+3-\dfrac{1}{y+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

\(P\le\dfrac{3\sqrt{2}}{2}-\dfrac{\sqrt{2}-1}{4}\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}\right)\)

Ta có:

\(\dfrac{1}{x+1}+\dfrac{1}{y+1}=\dfrac{x+y+2}{xy+x+y+1}=\dfrac{x+y+2}{x+y+5}=1-\dfrac{3}{x+y+5}\ge1-\dfrac{3}{2\sqrt{xy}+5}=\dfrac{2}{3}\)

\(\Rightarrow P\le\dfrac{3\sqrt{3}}{2}-\dfrac{\sqrt{2}-1}{4}.\dfrac{2}{3}=...\)

Dấu "=" xảy ra khi \(x=y=2\) hay \(\left(a;b\right)=\left(4;2\right)\)

Lời giải:

Áp dụng BĐT AM-GM:

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

\(\leq \frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{4(b+c)}+\frac{b}{b+a}+\frac{c}{4(c+b)}+\frac{c}{c+a}\)

\(=(\frac{a}{a+b}+\frac{b}{b+a})+(\frac{a}{a+c}+\frac{c}{a+c})+\frac{1}{4}(\frac{b}{b+c}+\frac{c}{b+c})=1+1+\frac{1}{4}=\frac{9}{4}\)

Vậy $P_{\max}=\frac{9}{4}$

Ta có: 

\(4\le\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=\sqrt{ab}+\sqrt{a}+\sqrt{b}+1\le\dfrac{a+b}{2}+\dfrac{a+1}{2}+\dfrac{b+1}{2}+1\)

\(=a+b+2\)

\(\Leftrightarrow a+b\ge2\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge\dfrac{\left(a+b\right)^2}{a+b}=a+b\ge2\)

Dấu \(=\) xảy ra khi \(a=b=1\).

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Ta đặt:

     \(\left\{{}\begin{matrix}x=a-1\\y=b-2\\z=c-3\end{matrix}\right.\)

        \(\Rightarrow x+y+z=3\) và  \(x,y,z\ge0\) (*)

Biểu thứ P trở thành:

     \(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\)

Từ (*) dễ thấy:

     \(\left\{{}\begin{matrix}0\le x\le3\\0\le y\le3\\0\le z\le3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le x\le\sqrt{3x}\\0\le y\le\sqrt{3y}\\0\le z\le\sqrt{3z}\end{matrix}\right.\)

Do đó:

     \(P\ge\dfrac{x+y+z}{\sqrt{3}}=\sqrt{3}\)

Dầu "=" xảy ra khi \(\left(a;b;c\right)=\left(3;0;0\right)=\left(0;3;0\right)=\left(0;0;3\right)\)