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Lời giải:
\(Q=\sqrt{a+b+c+2\sqrt{ab+bc}}+\sqrt{a+b+c+2\sqrt{ac+bc}}\)
\(=\sqrt{(a+c)+b+2\sqrt{b(a+c)}}+\sqrt{(a+b)+c+2\sqrt{c(a+b)}}\)
\(=\sqrt{(\sqrt{a+c}+\sqrt{b})^2}+\sqrt{(\sqrt{a+b}+\sqrt{c})^2}\)
\(=\sqrt{a+c}+\sqrt{b}+\sqrt{a+b}+\sqrt{c}\)
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 \(a+b< c\Rightarrow\sqrt{a+b}-\sqrt{c}< 0\)
\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{c}-\sqrt{a+b}=2\sqrt{c}\)
a)Áp dụng BĐT AM-GM ta có
\(\frac{ab\sqrt{ab}}{a+b}\le\frac{ab\sqrt{ab}}{2\sqrt{ab}}=\frac{ab}{2}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\frac{bc\sqrt{bc}}{b+c}\le\frac{bc}{2};\frac{ac\sqrt{ac}}{a+c}\le\frac{ac}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=Σ\frac{ab\sqrt{ab}}{a+b}\le\frac{ab+bc+ca}{2}=VP\)
Khi \(a=b=c\)
b)Áp dụng tiếp AM-GM:
\(b\sqrt{a-1}\le\frac{b\left(a-1+1\right)}{2}=\frac{ab}{2}\)
\(a\sqrt{b-1}\le\frac{a\left(b-1+1\right)}{2}=\frac{ab}{2}\)
Cộng theo vế 2 BĐT trên ta có:
\(VT=b\sqrt{a-1}+a\sqrt{b-1}\le ab=VP\)
Khi \(a=b=1\)
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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Cái này không khó :v
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+c}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Face khác ;v, theo AM-GM, ta có
\(\dfrac{a+b+c}{2}\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{6}{2}=3\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c=2
1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)
\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)
\(t^2=a+b+c+2\sqrt{ac+bc}+a+b+c-2\sqrt{ac+bc}+2\sqrt{\left(a+b+c+2\sqrt{ac+bc}\right)\left(a+b+c-2\sqrt{ac+bc}\right)}\)
\(T^2=2a+2b+2c+2\sqrt{a^2+b^2+c^2+2ab+2bc+2ac-4ac-4bc}\)
\(T^2=2a+2b+2c+\sqrt{a^2+b^2+c^2-2ac-2bc+2ab}\)
\(T^2=2a+2b+2c+\sqrt{\left(a+b-c\right)^2}\)
\(T^2=2a+2b+2c+a+b-c\) ( vì a,b,c> 0 )
\(T^2=3a+3b+c\Leftrightarrow t=\sqrt{3a+3b+c}\)