Ta có: $\sqrt[]{ab+2c}=\sqrt[]{ab+(a+b+c)c}=\sqrt[]{ab+ac+bc+c^2}=\sqrt[]{(c+a)(c+b)}$ (do $a+b+c=2$)

Nên $\dfrac{ab}{\sqrt[]{ab+2c}}=\dfrac{ab}{\sqrt[]{(c+a).(c+b)}}=ab.\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}}$

Áp dụng bất đẳng thức Cauchy cho $\dfrac{1}{a+c};\dfrac{1}{b+c}>0$ có:

$\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}} \leq \dfrac{1}{2}.(\dfrac{1}{a+c}+\dfrac{1}{b+c})$

Nên $\dfrac{ab}{\sqrt[]{ab+2c}} \leq \dfrac{1}{2}.ab.(\dfrac{1}{a+c}+\dfrac{1}{b+c})= \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})$

Tương tự ta có: $\dfrac{bc}{\sqrt[]{bc+2a}} \leq \dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$

$\dfrac{ca}{\sqrt[]{ca+2b}} \leq \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$

Nên $Q \leq  \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})+\dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})+ \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}.(\dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}=\dfrac{1}{2}.(a+b+c)=1$ (do $a+b+c=2$)

Dấu $=$ xảy ra khi $a=b=c=\dfrac{2}{3}$

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

GTNN của P là \(\dfrac{1}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)

\(Q=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)

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

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

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

\(=\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\)

\(a^2-ab+b^2=\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2\ge\dfrac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow P\le\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

để ý cái này: \(\sum\dfrac{a}{a+2\sqrt{bc}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}}=1\)

\(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)

\(P=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{ca}{b\left(a+b+c\right)+ca}}\)

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

\(P=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{a}{a+c}+\dfrac{b}{b+c}}{2}\\\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{b}{a+b}+\dfrac{c}{a+c}}{2}\\\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{c}{b+c}}{2}\end{matrix}\right.\)

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

\(\Rightarrow VT\le\dfrac{\dfrac{a+c}{a+c}+\dfrac{b+c}{b+c}+\dfrac{a+b}{a+b}}{2}=\dfrac{3}{2}\)

\(\Rightarrow P\le\dfrac{3}{2}\)

Vậy \(P_{max}=\dfrac{3}{2}\)

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

\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)

\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)

\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)

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

\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)

\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)

\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)

\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)

\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)