Từ giả thiết \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\Rightarrow xy+yz+xz=1\left(x=\dfrac{1}{a};y=\dfrac{1}{b};z=\dfrac{1}{c}\right)\)

\(A=\sum\dfrac{1}{\sqrt{1+a^2}}=\sum\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{a^2}+1}}=\sum\dfrac{x}{\sqrt{x^2+1}}=\sum\dfrac{x}{\sqrt{x^2+xy+yz+xz}}=\sum\dfrac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{1}{2}\sum\dfrac{x}{x+y}+\dfrac{x}{x+z}=\dfrac{3}{2}\)

Ta có: bc(a²+1) = (a+b)(a+c)

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

Áp dụng BĐT Cô-si: \(\sqrt{\dfrac{a}{a+b}}.\sqrt{\dfrac{a}{a+c}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{b+c}+\dfrac{a}{a+c}\right)\)

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

CMTT: \(\dfrac{b}{\sqrt{ac\left(1+b^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{a+c}\right)\)

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

\(\Rightarrow\) S \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{b+a}+\dfrac{a}{c+a}+\dfrac{b}{a+b}+\dfrac{b}{c+b}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)

\(\Rightarrow\) S\(\le\) \(\dfrac{1}{2}.3=\dfrac{3}{2}\)

Vậy S_max = \(\dfrac{3}{2}\)

Dấu "=" xảy ra\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=b=c\\a+b+c=abc\end{matrix}\right.\)

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

\(\dfrac{1}{\sqrt{a^2-ab+b^2}}< =\dfrac{1}{\sqrt{2ab-ab}}=\dfrac{1}{\sqrt{ab}}\)

\(\sqrt{\dfrac{1}{b^2-bc+c^2}}< =\dfrac{1}{\sqrt{bc}};\sqrt{\dfrac{1}{c^2-ac+c^2}}< =\dfrac{1}{\sqrt{ac}}\)

=>P<=1/a+1/b+1/c=3

Dấu = xảy ra khi a=b=c=1

Đặt \(\left(a;b;c\right)=\left(x^4;y^4;z^4\right)\Rightarrow xyz=1\)

\(VT=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(VT=\dfrac{1}{x^2+y^2+y^2+1+2}+\dfrac{1}{y^2+z^2+z^2+1+2}+\dfrac{1}{z^2+x^2+x^2+1+2}\)

\(VT\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)

\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)

\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)

\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)

Lời giải:
\(a+b+c=abc\)

\(\Rightarrow a(a+b+c)=a^2bc\)

\(\Rightarrow a(a+b+c)+bc=a^2bc+bc\)

\(\Rightarrow (a+b)(a+c)=bc(a^2+1)\)

\(\Rightarrow \frac{a}{\sqrt{bc(a^2+1)}}=\frac{a}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\) (theo BĐT AM-GM ngược dấu)

Hoàn toàn tương tự:

\(\frac{b}{\sqrt{ca(b^2+1)}}\leq \frac{1}{2}\left(\frac{b}{b+a}+\frac{b}{b+c}\right)\)

\(\frac{c}{\sqrt{ab(c^2+1)}}\leq \frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\)

Cộng theo vế những BĐT thu được ở trên ta có:

\(S\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)

Vậy \(S_{\max}=\frac{3}{2}\Leftrightarrow a=b=c=\sqrt{3}\)

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