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\(t^2=a+b+c+2\sqrt{ac+bc}+a+b+c-2\sqrt{ac+bc}+2\sqrt{\left(a+b+c+2\sqrt{ac+bc}\right)\left(a+b+c-2\sqrt{ac+bc}\right)}\)
\(T^2=2a+2b+2c+2\sqrt{a^2+b^2+c^2+2ab+2bc+2ac-4ac-4bc}\)
\(T^2=2a+2b+2c+\sqrt{a^2+b^2+c^2-2ac-2bc+2ab}\)
\(T^2=2a+2b+2c+\sqrt{\left(a+b-c\right)^2}\)
\(T^2=2a+2b+2c+a+b-c\) ( vì a,b,c> 0 )
\(T^2=3a+3b+c\Leftrightarrow t=\sqrt{3a+3b+c}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{ax}\frac{1}{\sqrt{a}}+\sqrt{by}\frac{1}{\sqrt{b}}+\sqrt{cz}\frac{1}{\sqrt{c}}\)
\(\le\sqrt{\left(ax+by+cz\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\sqrt{2S_{ABC}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\)
\(=\sqrt{\frac{abc}{2R}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\sqrt{\frac{ab+bc+ca}{2R}}\le\sqrt{\frac{a^2+b^2+c^2}{2R}}\)
2/
- Chứng minh \(\sqrt{2}\left(a+b+c\right)\le\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\)
Ta có \(\sqrt{2}.\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\sqrt{2}\left(a+b+c\right)\le\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\)
- Chứng minh \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}< \sqrt{3}\left(a+b+c\right)\)
Bạn chứng minh bằng biến đổi tương đương
1/ \(ab+bc+ac=3abc\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Ta có \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\le\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
Vậy min P = 3/2 tại a = b = c = 1
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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