Áp dụng BĐT cosi: 

\(\sqrt{\dfrac{b+c}{a}}\le\dfrac{\dfrac{b+c}{a}+1}{2}=\dfrac{\dfrac{a+b+c}{a}}{2}=\dfrac{a+b+c}{2a}\\ \Leftrightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

Cmtt \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{c+a}}\ge\dfrac{2c}{a+b+c}\)

Cộng vế theo vế 3 BĐT trên:

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

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c+a\\c=a+b\end{matrix}\right.\Leftrightarrow a+b+c=2\left(a+b+c\right)\)

\(\Leftrightarrow a+b+c=0\) (vô lí vì \(a,b,c>0\))

Do đó dấu "=" ko xảy ra hay \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\)

\(\dfrac{\dfrac{b+c}{a}+\dfrac{a}{a}}{2}>=\sqrt{\dfrac{b+c}{a}\cdot\dfrac{a}{a}}\)

=>\(\dfrac{a+b+c}{2a}>=\sqrt{\dfrac{b+c}{a}}\)

=>\(\sqrt{\dfrac{a}{b+c}}>=\dfrac{2a}{a+b+c}\)

Tương tự, ta có: \(\sqrt{\dfrac{b}{a+c}}>=\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}>=\dfrac{2c}{a+b+}\)

=>A>=2

Ace Legona

Bài 2:

\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)

Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)

\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:

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

Thiết lập các BĐT tương tự:

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

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)

Dấu "=" không xảy ra nên ta có ĐPCM

Lưu ý: lần sau đăng từng bài 1 thôi nhé !

1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)

TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)

\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)

Cộng vế với vế ta được:

\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)

Lời giải:

$\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$

$\Leftrightarrow a+b=a+c+b+c+2\sqrt{(a+c)(b+c)}$

$\Leftrightarrow 2c+2\sqrt{(a+c)(b+c)}=0$

$\Leftrightarrow c+\sqrt{(a+c)(b+c)}=0$

\(\Leftrightarrow \left\{\begin{matrix} -c=\sqrt{(a+c)(b+c)}\\ c< 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} c^2=(c+a)(c+b)\\ c< 0\end{matrix}\right.\)

\( \Leftrightarrow \left\{\begin{matrix} ab+bc+ac=0\\ c< 0\end{matrix}\right.\Leftrightarrow \frac{ba+bc+ac}{abc}=0\) (do $a,b>0$)

$\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

 (đpcm)

\(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)

\(\Leftrightarrow a+b=a+c+b+c+2\sqrt{\left(a+c\right)\left(b+c\right)}\)

\(\Leftrightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\)

\(\Leftrightarrow c+\sqrt{\left(a+c\right)\left(b+c\right)}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\-c=\sqrt{\left(a+c\right)\left(b+c\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\c^2=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c< 0\\ab+bc+ac=0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\left(đúng\right)\)

Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0

Khi đó:

(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)

=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2

=a+b+2c+2|c|=a+b+2c+2|c|

Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c

Do đó:

(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b

⇒√a+c+√b+c=√a+b