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Sửa đề: \(\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}\ge\sqrt{2}\left(\Sigma\sqrt{\frac{1-a}{a}}\right)\)
or \(\Sigma\frac{b+c}{a}\ge\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\)
Theo AM-GM:\(\frac{b+c}{a}\ge2\sqrt{\frac{2\left(b+c\right)}{a}}-2\)
Tương tự và cộng lại: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-6\)
Mà: \(\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\ge3\sqrt[6]{\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\ge6\)
Từ đó: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}=VP\)
Done!
\(\frac{a}{\sqrt{b}-1}+\frac{b}{\sqrt{c}-1}+\frac{c}{\sqrt{c}-1}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}=\frac{t^2}{t-3}=12.,\)
\(t^2-12t+36=0\Leftrightarrow t=6;.\)
=>a =b =c = 4
Ta có:
\(\frac{2}{\sqrt{a}}+\frac{2}{\sqrt{b}}+\frac{2}{\sqrt{c}}=\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)+\left(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)+\left(\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}\right)\)
\(\ge\frac{\left(1+1\right)^2}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+1\right)^2}{\sqrt{b}+\sqrt{c}}+\frac{\left(1+1\right)^2}{\sqrt{c}+\sqrt{a}}\)
\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{4}{\sqrt{b}+\sqrt{c}}+\frac{4}{\sqrt{c}+\sqrt{a}}\)
=> \(2\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)\(\ge4\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)
=> \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)\(\ge2\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)
"=" xảy ra <=> a =b =c.
ta có \(\sqrt{\frac{a}{1-a}}=\frac{a}{\sqrt{a\left(1-a\right)}}\)
áp dụng cô si
\(\sqrt{a\left(1-a\right)}< =\frac{a+1-a}{2}=\frac{1}{2}\)
do do\(\sqrt{\frac{a}{1-a}}>=2a\)\(\sqrt{\frac{b}{1-b}}>=2b,\sqrt{\frac{c}{1-c}}>=2c\)
cmtt\(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}>=2\left(a+b+c\right)=2\left(doa+b+c=1\right)\)
dau = xay ra <=>\(\hept{\begin{cases}a=1-a\\b=1-b\\c=1-c\end{cases}=>a+b+c=3-\left(a+b+c\right)}\)
<=>2(a+b+c)=3
<=>a+b+c=3/2
vay dau = khong xay ra ta co dpcm
Áp dụng bđt AM-GM:
\(\sum\sqrt{\frac{a}{1-a}}=\sum\frac{a}{\sqrt{a\left(b+c\right)}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu "=" không xảy ra =>đpcm
Ta có: \(4b\sqrt{c}-c\sqrt{a}=\sqrt{c}\left(4b-\sqrt{ac}\right)>0\)( do \(1< a,b,c< 2\))
Tương tự => Các MS dương
\(VT=\frac{ba}{4b\sqrt{ac}-ca}+\frac{cb}{4c\sqrt{ba}-ab}+\frac{ac}{4a\sqrt{bc}-bc}\)
Áp dụng BĐT cosi schawr ta có
\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{4b\sqrt{ac}+4c\sqrt{ab}+4a\sqrt{bc}-ab-bc-ac}\)
Áp dụng cosi ta có \(2b\sqrt{ac}=2\sqrt{ab}.\sqrt{ac}\le ab+ac\);\(2c\sqrt{ab}\le ac+bc\);\(2a\sqrt{bc}\le ab+ac\)
=> \(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{ab+bc+ac+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}}=\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}=1\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c