1/ \(Q=\frac{\left(2-\sqrt{a}\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}=2-\sqrt{a}\)

Do \(\sqrt{a}\ge0\Rightarrow2-\sqrt{a}\le2\Rightarrow Q_{max}=2\) khi \(a=0\)

2/

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

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

\(=\sqrt{a+b}+\sqrt{c}+\left|\sqrt{a+b}-\sqrt{c}\right|\)

TH1: Nếu \(a+b\ge c\Rightarrow\sqrt{a+b}-\sqrt{c}\ge0\)

\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{a+b}-\sqrt{c}=2\sqrt{a+b}\)

TH2: Nếu \(a+b< c\Rightarrow\sqrt{a+b}-\sqrt{c}< 0\)

\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{c}-\sqrt{a+b}=2\sqrt{c}\)

Ta có : \(\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}=\sqrt{a+b+2\sqrt{c}.\sqrt{a+b}+c}+\sqrt{a+b-2\sqrt{c}.\sqrt{a+b}+c}=\sqrt{\left(\sqrt{a+b}+\sqrt{c}\right)^2}+\sqrt{\left(\sqrt{a+b}-\sqrt{c}\right)^2}\)\(=\sqrt{a+b}+\sqrt{c}+\left|\sqrt{a+b}-\sqrt{c}\right|=\sqrt{a+b}+\sqrt{c}+\left(\sqrt{a+b}-\sqrt{c}\right)=2\sqrt{a+b}\)(vì a,b,c là độ dài ba cạnh của tam giác nên \(a+b>c>0\Rightarrow\sqrt{a+b}>\sqrt{c}\))

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Đặt \(A=\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}\)
=> \(A^2=\left(\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}\right)^2\)
=>\(A^2=\left(a+b+c+2\sqrt{ab+bc}\right)+2\cdot\sqrt{a+b+c+2\sqrt{ab+bc}}\cdot\sqrt{a+b+c-2\sqrt{ab+bc}}+\left(a+b+c-2\sqrt{ab+bc}\right)\)
=>\(A^2=2a+2b+2c+2\cdot\sqrt{\left(\left(a+b+c\right)+2\sqrt{ab+bc}\right)\cdot\left(\left(a+b+c\right)-2\sqrt{ab+bc}\right)}\)
=>\(A^2=2a+2b+2c+2\cdot\sqrt{\left(a+b+c\right)^2-4ac-4bc}\)
=>\(A^2=2a+2b+2c+2\cdot\sqrt{\left(a+b-c\right)^2}\)
=>\(A^2=2a+2b+2c+2a+2b-2c=4a+4b=4\left(a+b\right)\)
=>\(A=\sqrt{A^2}=\sqrt{4\left(a+b\right)}=2\sqrt{a+b}\)

Lời giải:

\(Q=\sqrt{a+b+c+2\sqrt{ab+bc}}+\sqrt{a+b+c+2\sqrt{ac+bc}}\)

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

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

\(=\sqrt{a+c}+\sqrt{b}+\sqrt{a+b}+\sqrt{c}\)

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

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

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

Tương tự cho 2 đẳng thức còn lại rồi cộng theo vế

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

ta có : \(P=\frac{\sqrt{bc}}{a+2\sqrt{bc}}+\frac{\sqrt{ac}}{b+2\sqrt{ac}}+\frac{\sqrt{ab}}{c+2\sqrt{ab}}\le\frac{\frac{1}{2}\left(b+c\right)}{a+b+c}+\frac{\frac{1}{2}\left(a+c\right)}{a+b+c}+\frac{\frac{1}{2}\left(a+b\right)}{a+b+c}\)

\(\Rightarrow P\le\frac{a+b+c}{a+b+c}=1\)

=> GTLN của P là 1 khi a=b=c

Áp dụng BĐT Cô-si:

\(A\le\dfrac{a+b}{2\sqrt{c+ab}}+\dfrac{b+c}{2\sqrt{a+bc}}+\dfrac{c+a}{2\sqrt{b+ac}}\)\(\le\dfrac{a+b}{2\sqrt{2\sqrt{abc}}}+\dfrac{b+c}{2\sqrt{2\sqrt{abc}}}+\dfrac{c+a}{2\sqrt{2\sqrt{abc}}}\)\(=\dfrac{a+b+c}{\sqrt[4]{4abc}}=\dfrac{1}{\sqrt[4]{4abc}}\ge\dfrac{1}{\sqrt{\left(a+b+c\right).\dfrac{2}{3}}}\)(BĐT Cô-si)\(=\dfrac{1}{\sqrt{\dfrac{2}{3}}}=\dfrac{\sqrt{6}}{2}\)

Vậy A_min=\(\dfrac{\sqrt{6}}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)

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

Tương tự

=> P \(\ge\dfrac{\sqrt{3}}{2}.2\left(a+b+c\right)=\sqrt{3}.\)

Vậy \(Pmin=\sqrt{3}\) khi a =b=c = 1/3