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

\(T^2=2a+2b+2c+2\sqrt{a^2+b^2+c^2+2ab+2bc+2ac-4ac-4bc}\)

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

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

\(T^2=2a+2b+2c+a+b-c\) ( vì a,b,c> 0 )

\(T^2=3a+3b+c\Leftrightarrow t=\sqrt{3a+3b+c}\)

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

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

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

\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)

\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự:

\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)

\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)

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

Đặt \(x=\sqrt{bc};y=\sqrt{ca};z=\sqrt{ab}\)\(\Rightarrow x^2+y^2+z^2+xyz=4\)\(\Rightarrow\left(x+y+z\right)^2-4=2\left(xy+yz+zx\right)-xyz\)

\(\Rightarrow\left(x+y+z\right)^2-4\left(x+y-z\right)+4=\left(2-x\right)\left(2-y\right)\left(2-z\right)\)\(\le\left(\frac{6-x-y-z}{3}\right)^3\)

Đặt \(t=x+y+z\Rightarrow\left(t-6\right)^3+27\left(t^2-4t+4\right)\le0\)\(\Leftrightarrow\left(t-3\right)\left(t+6\right)^2\le0\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\left(đpcm\right)\)

Dấu '=' xảy ra <=> a=b=c=1

Mình chưa hiểu ở dòng thứ 3 tại sao bạn lại đánh giá đc nó nhỏ hơn hoặc bằng \(\left(\frac{6-x-y-z}{3}\right)^3\)

\(P=\dfrac{1}{2}\left(\dfrac{2\sqrt{bc}}{a+2\sqrt{bc}}+\dfrac{2\sqrt{ac}}{b+2\sqrt{ac}}+\dfrac{2\sqrt{ab}}{c+2\sqrt{ab}}\right)\)

\(P=\dfrac{1}{2}\left(1-\dfrac{a}{a+2\sqrt{bc}}+1-\dfrac{b}{b+2\sqrt{ca}}+1-\dfrac{c}{c+2\sqrt{ab}}\right)\)

\(P=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a}{a+2\sqrt{bc}}+\dfrac{b}{b+2\sqrt{ca}}+\dfrac{c}{c+2\sqrt{ab}}\right)\)

\(P\le\dfrac{3}{2}-\dfrac{1}{2}.\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+2\sqrt{bc}+b+2\sqrt{ca}+c+2\sqrt{ab}}=\dfrac{3}{2}-\dfrac{1}{2}.\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}=1\)

\(P_{max}=1\) khi \(a=b=c\)

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

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

\(\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}\)

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

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

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

Đến đây e ko bít làm tiếp -_- 

Chúc chị học tốt ~ 

a/ Nếu (a + b) < 0 thì bất  đẳng thức đúng

Với (a + b) \(\ge0\)thì ta có

\(2a^2+ab+2b^2\ge\frac{5}{4}\left(a^2+2ab+b^2\right)\)

\(\Leftrightarrow3a^2-6ab+3b^2\ge0\)

\(\Leftrightarrow3\left(a-b\right)^2\ge0\)(đúng)

b/ Áp dụng BĐT BCS : 

\(1=\left(1.\sqrt{a}+1.\sqrt{b}+1.\sqrt{c}\right)^2\le3\left(a+b+c\right)\Rightarrow a+b+c\ge\frac{1}{3}\)

Áp dụng câu a/ :

\(\sqrt{2a^2+ab+2b^2}\ge\frac{\sqrt{5}}{2}\left(a+b\right)\)

\(\sqrt{2b^2+bc+2c^2}\ge\frac{\sqrt{5}}{2}\left(b+c\right)\)

\(\sqrt{2c^2+ac+2a^2}\ge\frac{\sqrt{5}}{2}\left(a+c\right)\)

\(\Rightarrow P\ge\frac{\sqrt{5}}{2}.2\left(a+b+c\right)\ge\frac{\sqrt{5}}{3}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{9}\)

Vậy min P = \(\frac{\sqrt{5}}{3}\) khi a=b=c=1/9

Lời giải:

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

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

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

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