Theo BĐT Cauchy : 

\(\sqrt{\frac{b+c}{a}.1}\le\frac{\frac{b+c}{a}+1}{2}=\frac{a+b+c}{2a}\)

Do đó : \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\)

Tương tự : \(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\)

              \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)

\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Dấu "=" xảy ra khi và chỉ khi :

\(\hept{\begin{cases}a=b+c\\b=c+a\\c=a+b\end{cases}\Rightarrow a+b+c=0}\), vô lí vì a, b, c là các số dương nên đẳng thức không xảy ra.

Vậy \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\).

Chết cha, mình bị thiếu chỗ dấu "=" xảy ra là c = a + b.

\(2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\)

Thay thế \(a+b+c=1\)

\(\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a+b+c}{b+c}+\frac{a+2b+c}{a+c}+\frac{a+b+2c}{a+b}\)

\(\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3\)

\(\Leftrightarrow\frac{2b}{a}+\frac{2c}{b}+\frac{2a}{c}\ge\frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}+3\)

\(\Leftrightarrow\left(\frac{2b}{a}-\frac{2b}{a+c}\right)+\left(\frac{2c}{b}-\frac{2c}{a+b}\right)+\left(\frac{2a}{c}-\frac{2a}{b+c}\right)\ge3\)

\(\Leftrightarrow\frac{2bc}{a\left(a+c\right)}+\frac{2ca}{b\left(a+b\right)}+\frac{2ab}{c\left(b+c\right)}\ge3\)

\(\Leftrightarrow\frac{bc}{a\left(a+c\right)}+\frac{ca}{b\left(a+b\right)}+\frac{ab}{c\left(b+c\right)}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}\)

Áp dụng bất đẳng thức cộng mẫu số 

\(\Rightarrow\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\)

\(\ge\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c+a+b+c\right)}=\frac{\left(ab+bc+ca\right)^2}{2abc}\)

Chứng minh rằng : \(\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}\)

\(\Leftrightarrow2\left(ab+bc+ca\right)^2\ge6abc\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\ge3abc\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\hept{\begin{cases}a^2b^2+b^2c^2\ge2\sqrt{a^2b^4c^2}=2ab^2c\\b^2c^2+c^2a^2\ge2\sqrt{a^2b^2c^4}=2abc^2\\a^2b^2+c^2a^2\ge2\sqrt{a^2b^2c^2}=2a^2bc\end{cases}}\)

\(\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\left(đpcm\right)\)

Vì \(\frac{\left(ab+bc+ca\right)^2}{2abc}\ge\frac{3}{2}\)

Vậy \(\frac{\left(bc\right)^2}{abc\left(a+c\right)}+\frac{\left(ca\right)^2}{abc\left(a+b\right)}+\frac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\frac{3}{2}\)

\(\Leftrightarrow2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\left(đpcm\right)\)

Chúc bạn học tốt !!!

Áp dụng bđt Cô si với 2 số dương là: \(\sqrt{\frac{b+c}{a}}\) và 1 ta có:

\(\left(\frac{b+c}{a}+1\right):2\ge\sqrt{\frac{b+c}{a}.1}\)

\(\Leftrightarrow\) \(\frac{a+b+c}{2a}\ge\sqrt{\frac{b+c}{a}}\)

hay \(\sqrt{\frac{a}{b+c}}\ge\frac{2a}{a+b+c}\left(1\right)\)

Tương tự như trên ta cũng có:

\(\sqrt{\frac{b}{a+c}}\ge\frac{2b}{a+b+c}\left(2\right)\)

\(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\left(3\right)\)

Từ (1); (2) và (3) \(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=\frac{2.\left(a+b+c\right)}{a+b+c}=2\)

Dấu "=" xảy ra khi \(\begin{cases}\sqrt{\frac{b+c}{a}}=1\\\sqrt{\frac{a+c}{b}}=1\\\sqrt{\frac{a+b}{c}}=1\end{cases}\)\(\Leftrightarrow\begin{cases}\frac{b+c}{a}=1\\\frac{a+c}{b}=1\\\frac{a+b}{c}=1\end{cases}\)\(\Leftrightarrow\begin{cases}b+c=a\\a+c=b\\a+b=c\end{cases}\)

\(\Rightarrow2.\left(a+b+c\right)=a+b+c\)\(\Rightarrow a+b+c=0\), mâu thuẫn với đề bài a; b; c là các số dương

Như vậy dấu "=" không xảy ra

Do đó, \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\left(đpcm\right)\)

Hic .

Định làm mak bj cúp điện nên nghỉ luôn

Ai ngờ dk vào câu hỏi hay

<xxx>

\(\sqrt{\frac{a}{b+c}}=\frac{a}{a\sqrt{b+c}}\ge\frac{2a}{a+b+c}\)

Tương tự: \(\sqrt{\frac{b}{c+a}}\ge\frac{2b}{a+b+c}\) ; \(\sqrt{\frac{c}{a+b}}\ge\frac{2c}{a+b+c}\)

\(\Rightarrow\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Dấu "=" ko xảy ra nên \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\)

ra nguyên những bài khó hiểu hết là chết liền

hoc lop may vay bai kho qua ai ma giai duoc

Ta có ; \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)(1)

Nhân cả hai vế của (1) với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được : 

\(\Leftrightarrow\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)^2}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}+\frac{c}{\left(a-b\right)^2}=0\)

\(\Leftrightarrow\left[\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\right]+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(a-b\right)\left(c-a\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)\(\Leftrightarrow\left[\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\right]+\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)

Vậy ta có điều phải chứng minh.