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

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

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

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

_ Chứng minh VT <2 .

Với a,b,c > 0, ta có:

\(a< a+b\Rightarrow\dfrac{a}{a+b}< 1=\dfrac{c}{c}\Rightarrow\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\) (1)

\(b< b+c\Rightarrow\dfrac{b}{b+c}< 1=\dfrac{a}{a}\Rightarrow\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c}\) (2)

\(c< c+a\Rightarrow\dfrac{c}{c+a}< 1=\dfrac{b}{b}\Rightarrow\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\) (3)

Từ (1) , (2) và (3), Cộng vế theo vế ta có:

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\)(*)

_Chứng minh VP > 2.

Theo BĐT Cô-si, ta có:

\(\sqrt{\dfrac{b+c}{a}.1}\le\left(\dfrac{b+c}{a}+1\right):2=\dfrac{b+c+a}{2a}\)

Do vậy : \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

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

Cộng vế theo vế

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

Dấu ''='' xảy ra \(\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\)

\(\Rightarrow a+b+c=0\) (trái với g/t a,b,c >0)

Vậy đẳng thức khong xảy ra dấu ''=''

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

Từ (*) và (**) \(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\)

Lời giải:

Đặt \(\left ( \frac{\sqrt{a^2+b^2}}{c},\frac{\sqrt{b^2+c^2}}{a}, \frac{\sqrt{c^2+a^2}}{b} \right )=(x,y,z)\)

BĐT cần chứng minh tương đương với:
\(x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)\((*)\)

------------------------------------------------------------------

Từ cách đặt $x,y,z$ ta có:

\(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=1\)

Áp dụng BĐT Bunhiacopxky:

\(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}=\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\left(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}\right)\)

\(\geq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)

\(\Leftrightarrow 3\geq 2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

\(\Leftrightarrow xyz\geq \frac{2}{3}(x+y+z)\)

\(\Rightarrow xyz(x+y+z)\geq \frac{2}{3}(x+y+z)^2\)

Áp dụng BĐT AM_GM ta lại có:

\((x+y+z)^2\geq 3(xy+yz+xz)\). Do đó:

\(xyz(x+y+z)\geq 2(xy+yz+xz)\)

\(\Leftrightarrow x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Đúng theo \((*)\)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

áp dụng bat dang thuc bunhiacóki

ta có \(\dfrac{\sqrt{a^2+b^2}}{c}\ge\dfrac{a+b}{\sqrt{2}c}\)

ttu vt \(\ge\dfrac{1}{\sqrt{2}}\left(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\right)\)

=\(\dfrac{a}{\sqrt{2}}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{b}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+\dfrac{c}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) (1)

áp dung bdt \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

ta có (1) \(\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\)

tiếp tục áp dụng bunhia ta có \(\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{2a}{\sqrt{b^2+c^2}}\)

ttuong tu ta có \(vt\ge2\left(\dfrac{a}{\sqrt{b^2+c2}}+\dfrac{b}{\sqrt{a^2+c^2}}+\dfrac{c}{\sqrt{a^2+b^2}}\right)\left(dpcm\right)\)