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.

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

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

sửa đề lại bạn nhé =) \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)

đặt \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=k\Rightarrow\hept{\begin{cases}a=kA\\b=kB\end{cases}va\hept{\begin{cases}c=kC\\d=kD\end{cases}}}\)

theo đề bài ta có \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{kA^2}+\sqrt{kB^2}+\sqrt{kC^2}+\sqrt{kD^2}\)

=\(\sqrt{k}\left(A+B+C+D\right)\left(1\right)\)

ta lại có \(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{\left(kA+kB+kC+kD\right)\left(A+B+C+D\right)}\)

=\(\sqrt{k\left(A+B+C+D\right)\left(A+B+C+D\right)}=\sqrt{k\left(A+B+C+D\right)^2}=\sqrt{k}\left(A+B+C+D\right)\left(2\right)\)

(1),(2)=> \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)

gt thiếu kìa. 

Đặt \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=k\left(k>0\right)\)

\(\Rightarrow a=Ak,b=Bk,c=Ck,d=Dk\)

Ta có: 

* \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}\)

\(=\sqrt{kA^2}+\sqrt{kB^2}+\sqrt{kC^2}+\sqrt{kD^2}\)

\(=\left(A+B+C+D\right).\sqrt{k}\left(1\right)\)

* \(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)

\(=\sqrt{k.\left(A+B+C+D\right)^2}\)

\(=\sqrt{k}.\left(A+B+C+D\right)\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)

(Chúc bạn học tốt và k cho mình với nhé!)

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

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2019}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\) \(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)

\(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{4\left(x+y+z\right)^2}{2x+2y+2z}-\left(x+y+z\right)=x+y+z=\sqrt{2019}\)

\(\Rightarrow VT\ge\dfrac{\sqrt{2019}}{2\sqrt{2}}=\sqrt{\dfrac{2019}{8}}\) (đpcm)

\(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=\frac{a+b+c+d}{A+B+C+D}\)

\(\Rightarrow A.a=\frac{A^2\left(a+b+c+d\right)}{A+B+C+D}\Rightarrow\sqrt{Aa}=\frac{A\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\)

Tương tự ta có: \(\sqrt{Bb}=\frac{B\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\); \(\sqrt{Cc}=\frac{C\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\); \(\sqrt{Dd}=\frac{D\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\)

Cộng vế với vế:

\(\sqrt{Aa}+\sqrt{Bb}+\sqrt{Cc}+\sqrt{Dd}=\frac{\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\left(A+B+C+D\right)=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)

Làm cách này chắt đuoc

Ap dung BDT Bun-nhi-a-cop-xki ta co:

\(\left(\sqrt{Aa}+\sqrt{Bb}+\sqrt{Cc}+\sqrt{Dd}\right)^2\le\left(A+B+C+D\right)\left(a+b+c+d\right)\)

\(\Rightarrow\sqrt{Aa}+\sqrt{Bb}+\sqrt{Cc}+\sqrt{Dd}\le\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)Dau '=' xay ra khi \(\frac{A}{a}=\frac{B}{b}=\frac{C}{c}=\frac{D}{d}\)hay \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)

Ma theo gia thuyet cua de bai thi:

\(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)

Nen dang thuc tren ton tai voi \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)

sử dụng bdt bunhiacopxki có đc ko bn

\(a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)

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

\(\ge2a+2b+2c\ge6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\)

Đặt đẳng thức là A. Áp dụng bất đẳng thức AM-GM ta có:

\(\sqrt{2b\left(a-b\right)}\le\frac{2b+\left(a+b\right)}{2}=\frac{a+3b}{2}\)

Từ đó: \(A\ge\frac{2a\sqrt{2}}{a+3b}+\frac{2b\sqrt{2}}{b+3c}+\frac{2c\sqrt{2}}{c+3a}\)

Ta sẽ chứng minh: \(M=\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)

Thật vậy, ta có: \(M=\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ca}\)

Theo BĐT AM-GM ta có:

\(ab+bc+ca\le a^2+b^2+c^2\)

Áp dụng BĐT cauchy ta được:

\(M\ge\frac{\left(a+b+c\right)^2}{\frac{4}{3}\left(a^2+b^2+c^2\right)+\frac{8}{3}\left(ab+bc+ca\right)}\)\(=\frac{\left(a+b+c\right)^2}{\frac{4}{3}\left(a+b+c\right)^2}=\frac{3}{4}\)

Vì vậy: \(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)

Từ đó ta có: \(A\ge\frac{2a\sqrt{2}}{a+3b}+\frac{2b\sqrt{2}}{b+3c}+\frac{2c\sqrt{2}}{c+3a}\ge2\sqrt{2}.\frac{3}{4}=\frac{3\sqrt{2}}{2}\)

Vậy đẳng thức xảy xa khi và chỉ khi a=b=c