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bđt \(\Leftrightarrow\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)
Ta có: \(\left(\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\right)+\left(\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{bc}}{\sqrt{a}}\right)+\left(\dfrac{\sqrt{ab}}{\sqrt{c}}+\dfrac{\sqrt{ca}}{\sqrt{b}}\right)\ge2\sqrt{b}+2\sqrt{c}+2\sqrt{a}\)
\(\Leftrightarrow2\left(\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\Leftrightarrow\dfrac{\sqrt{bc}}{\sqrt{a}}+\dfrac{\sqrt{ca}}{\sqrt{b}}+\dfrac{\sqrt{ab}}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\left(đpcm\right)\)
1. Đặt $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=T$
$\frac{a}{b+c}> \frac{a}{a+b+c}$
$\frac{b}{c+a}> \frac{b}{c+a+b}$
$\frac{c}{a+b}> \frac{c}{a+b+c}$
$\Rightarrow T> \frac{a+b+c}{a+b+c}=1$ (đpcm)
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Xét hiệu:
$\frac{a}{b+c}-\frac{2a}{a+b+c}=\frac{-a(b+c-a)}{(b+c)(a+b+c)}<0$ theo BĐT tam giác
$\Rightarrow \frac{a}{b+c}< \frac{2a}{a+b+c}$
Tương tư: $\frac{b}{c+a}< \frac{2b}{c+a+b}$
$\frac{c}{a+b}< \frac{2c}{a+b+c}$
Cộng theo vế:
$T< \frac{2(a+b+c)}{a+b+c}=2$
$\frac{b}{a+c}
2.
Áp dụng BĐT AM-GM:
\(\frac{b+c}{a}.1\leq \frac{1}{4}(\frac{b+c}{a}+1)^2=\frac{(b+c+a)^2}{4a^2}\)
\(\Rightarrow \sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}\)
Tương tự với các phân thức còn lại và cộng theo vế:
$\Rightarrow T\geq \frac{2(a+b+c)}{a+b+c}=2$
Dấu "=" xảy ra khi $b+c=a; c+a=b; a+b=c\Rightarrow a=b=c=0$ (vô lý)
Vậy dấu "=" không xảy ra, tức là $T>2>1$ (đpcm)
\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)
Làm tương tự, cộng lại và rút gọn
Ta có: \(abc=b+2c\)
\(\Rightarrow a=\dfrac{b+2c}{bc}\)\(\Rightarrow a=\dfrac{1}{c}+\dfrac{2}{b}\)
Áp dụng bất đẳng thức: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
Ta có: \(\dfrac{3}{b+c-a}+\dfrac{4}{c+a-b}+\dfrac{5}{a+b-c}\)
\(=\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}+2\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+b-c}\right)+3\left(\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\right)\ge\dfrac{4}{b+c-a+c+a-b}+2.\dfrac{4}{b+c-a+a+b-c}+3.\dfrac{4}{c+a-b+a+b-c}=\dfrac{4}{2c}+2.\dfrac{4}{2b}+3.\dfrac{4}{2a}=\dfrac{2}{c}+\dfrac{4}{b}+\dfrac{6}{a}=2\left(\dfrac{1}{c}+\dfrac{2}{b}+\dfrac{3}{a}\right)=2\left(a+\dfrac{3}{a}\right)\ge2.2\sqrt{\dfrac{a.3}{a}}=4\sqrt{3}\)
(bất đẳng thức Cauchy cho 2 số dương)
\(ĐTXR\Leftrightarrow a=b=c=\sqrt{3}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
Áp dụng BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(\text{luôn đúng}\right)\)
Ta có \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Cộng VTV \(\Leftrightarrow3\ge\dfrac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\sqrt[3]{abc}+1\)
\(\Leftrightarrow VT^2=\sum\left[\dfrac{1}{a\left(b+1\right)}\right]^2\ge3\cdot\sum\dfrac{1}{ab\left(a+1\right)\left(b+1\right)}\\ \Leftrightarrow VT^2\ge3\cdot\dfrac{a^2+b^2+c^2+a+b+c}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\cdot\dfrac{a+b+c+ab+bc+ca}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\\ \Leftrightarrow VT^2\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(1+\sqrt[3]{abc}\right)^3}\\ \Leftrightarrow VT^2\ge\dfrac{9}{\sqrt[3]{\left(abc\right)^2}\left(1+\sqrt[3]{abc}\right)^2}=VP^2\\ \LeftrightarrowĐpcm\)
Dấu \("="\Leftrightarrow a=b=c=1\)