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Không làm mất tính tổng quát của bài toán, giả sử \(a\ge b\ge c\)(1)
Có \(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}=\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
Từ (1) => \(\hept{\begin{cases}\frac{2}{a}\le\frac{1}{a}+\frac{1}{b}\\\frac{2}{b}\le\frac{1}{b}+\frac{1}{c}\\\frac{2}{c}\le\frac{1}{a}+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{\frac{2}{a}}\le\sqrt{\frac{1}{a}+\frac{1}{b}}\\\sqrt{\frac{2}{b}}\le\sqrt{\frac{1}{b}+\frac{1}{c}}\\\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{a}+\frac{1}{c}}\end{cases}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)
Ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow\frac{a^2+2ab+b^2}{4}-ab\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng \(\forall a,b\) )
=>đpcm
Cô si
\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2c\)
\(\frac{ca}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}=2a\)
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)
Cộng lại ta có:
\(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(c+a+b\right)}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Rightarrow dpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> \(a+b\ge2\sqrt{ab}\)
<=> \(a+b-2\sqrt{ab}\ge0\)
<=. \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng )
dấu = khi a=b
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow a+b\ge2\sqrt{ab}\)
<=>\(a+b-2\sqrt{ab}\ge0\)
<=>\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(luôn đúng)
=>dpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhìn giả thiết thấy nản quả:(
BĐT \(\Leftrightarrow\Sigma_{cyc}\frac{\left(ab+bc+ca\right)\left(a+b\right)}{a^2+b^2}\le3\left(ab+bc+ca\right)\) (nhân ab +bc +ca vào hai vế)
\(\Leftrightarrow\Sigma_{cyc}\frac{\left(ab+bc+ca\right)\left(a+b\right)}{a^2+b^2}\le3\left(a+b+c\right)\) (chú ý giả thiết ab + bc +ca = a + b + c)
\(VT=\Sigma_{cyc}\frac{ab\left(a+b\right)}{a^2+b^2}+\Sigma_{cyc}\frac{c\left(a+b\right)^2}{a^2+b^2}\)
\(\le\Sigma_{cyc}\frac{ab\left(a+b\right)}{2ab}+\Sigma_{cyc}\frac{2c\left(a^2+b^2\right)}{a^2+b^2}=3\left(a+b+c\right)\)
Vậy ta có đpcm.Đẳng thức xảy ra khi a = b = c
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{\sqrt{AB}}=\frac{2}{2\sqrt{AB}}\ge\frac{2}{A+B}\)(đpcm)
p/s: tham khảo
chúc bn hk tốt