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\(a+b+c=abc\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\sqrt{1+\frac{1}{a^2}}+\sqrt{1+\frac{1}{b^2}}+\sqrt{1+\frac{1}{c^2}}\ge\sqrt{\left(1+1+1\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(\ge\sqrt{3^2+3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c=\(\sqrt{3}\)
\(\frac{a}{\sqrt{b}-1}+\frac{b}{\sqrt{c}-1}+\frac{c}{\sqrt{c}-1}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}=\frac{t^2}{t-3}=12.,\)
\(t^2-12t+36=0\Leftrightarrow t=6;.\)
=>a =b =c = 4
a) \(\sqrt{a}+1>\sqrt{a+1}\)\(\Leftrightarrow\)\(a+2\sqrt{a}+1>a+1\)\(\Leftrightarrow\)\(2\sqrt{a}>0\)( luôn đúng \(\forall x>0\) )
b) \(a-1< a\)\(\Leftrightarrow\)\(\sqrt{a-1}< \sqrt{a}\)
c) \(\left(\sqrt{6}-1\right)^2=6-2\sqrt{6}+1>3-2\sqrt{3.2}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
do \(\sqrt{6}-1>0;\sqrt{3}-\sqrt{2}>0\) nên \(\sqrt{6}-1>\sqrt{3}-\sqrt{2}\) ( đpcm )
\(\sqrt{\frac{a}{c+b}}=\frac{a}{\sqrt{a\left(c+b\right)}}\ge\frac{a}{\frac{a+b+c}{2}}=\frac{2a}{a+b+c}\)
tương tự : \(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c};\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)
\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)(ĐPCM)
\(\sum\sqrt{\frac{a}{1-a}}=\sum\frac{a}{\sqrt{a\left(b+c\right)}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\) (AM-GM)
Dấu "=" không xảy ra =>đpcm