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\(ab+bc+ca\le1\)
\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)
\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)
\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)
Ta có \(\sqrt{a-1}+\dfrac{1}{\sqrt{a-1}}\) \(=\sqrt{a-1}+\dfrac{1}{4\sqrt{a-1}}+\dfrac{3}{4\sqrt{a-1}}\) \(\ge2\sqrt{\sqrt{a-1}.\dfrac{1}{4\sqrt{a-1}}}+\dfrac{3}{4\sqrt{a-1}}\) \(=1+\dfrac{3}{4\sqrt{a-1}}\).
Lập 2 BĐT tương tự rồi cộng vế theo vế, ta có
\(VT\ge3+\dfrac{3}{4}\left(\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\right)\)
\(\ge3+\dfrac{3}{4}.\dfrac{9}{\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}}\)
\(\ge3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}\) \(=\dfrac{15}{2}\).
ĐTXR \(\Leftrightarrow a=b=c=\dfrac{5}{4}\). Ta có đpcm
Có \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{15}{2}-\left(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\right)\ge6\) (1)
Ta chứng minh (1) đúng
Áp dụng bất đẳng thức Schwarz :
\(\dfrac{1}{\sqrt{a-1}}+\dfrac{1}{\sqrt{b-1}}+\dfrac{1}{\sqrt{c-1}}\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}}\ge\dfrac{9}{\dfrac{3}{2}}=6\)Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{a-1}=\sqrt{b-1}=\sqrt{c-1}\\\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=\dfrac{5}{4}\)(tm)
\(\sum\dfrac{a}{\sqrt{ab+b^2}}=\sum\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\sum\dfrac{2\sqrt{2}a}{2b+a+b}=2\sqrt{2}\sum\dfrac{a}{a+3b}\)
\(=2\sqrt{2}\sum\dfrac{a^2}{a^2+3ab}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\)
\(=\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3\sqrt{2}}{2}\)
\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)
Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)
Cộng vế với vế:
\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)
Đề có thiếu gì không bạn?