Ta có: \(\sqrt{\frac{a}{b+c+d}}=\sqrt{\frac{a^2}{a\left(b+c+d\right)}}=\frac{a}{\sqrt{a\left(b+c+d\right)}}\)

Xét \(\sqrt{a\left(b+c+d\right)}\le\frac{a+b+c+d}{2}\)

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

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

(a,b,c,d>0)

Cmtt: \(\hept{\begin{cases}\sqrt{\frac{b}{a+c+d}}\ge\frac{2b}{a+b+c+d}\\\sqrt{\frac{c}{b+a+d}}\ge\frac{2c}{a+b+c+d}\\\sqrt{\frac{d}{a+b+c}}\ge\frac{2d}{a+b+c+d}\end{cases}}\)

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

Đến đây tự xử lí phần dấu "="

\(VT^2\ge\left(1+1+1+1\right)\left(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{d+a+b}+\frac{d}{b+a+c}\right)\ge4.1=4\)

=> VT >/ 2

Dễ CM được \(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{d+a+b}+\frac{d}{b+a+c}\ge1\)

\(\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\)

\(=\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\)

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

Dấu '' = '' xảy ra khi a = b + c+ d 

                              b = c+d+a 

                            c = b+a+d

                             d = a+b+c 

Hình như ko có a ; b; c ;d 

Xét: \(\sqrt{\frac{a}{b+c+d}}=\frac{\sqrt{a}}{\sqrt{b+c+d}}=\frac{a}{\sqrt{a\left(b+c+d\right)}}\)

\(\sqrt{\frac{b}{c+d+a}}=\frac{\sqrt{b}}{\sqrt{c+d+a}}=\frac{b}{\sqrt{b\left(c+d+a\right)}}\)

\(\sqrt{\frac{c}{d+a+b}}=\frac{\sqrt{c}}{\sqrt{d+a+b}}=\frac{c}{\sqrt{c\left(d+a+b\right)}}\)

\(\sqrt{\frac{d}{a+b+c}}=\frac{\sqrt{d}}{\sqrt{a+b+c}}=\frac{d}{\sqrt{d\left(a+b+c\right)}}\)

\(\Rightarrow VT=\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\)

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

\(\Rightarrow\left\{\begin{matrix}\sqrt{a\left(b+c+d\right)}\le\frac{a+b+c+d}{2}\\\sqrt{b\left(c+d+a\right)}\le\frac{a+b+c+d}{2}\\\sqrt{c\left(d+a+b\right)}\le\frac{a+b+c+d}{2}\\\sqrt{d\left(a+b+c\right)}\le\frac{a+b+c+d}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a\left(b+c+d\right)}}\ge\frac{2a}{a+b+c+d}\\\frac{b}{\sqrt{b\left(c+d+a\right)}}\ge\frac{2b}{a+b+c+d}\\\frac{c}{\sqrt{c\left(d+a+b\right)}}\ge\frac{2c}{a+b+c+d}\\\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge\frac{2d}{a+b+c+d}\end{matrix}\right.\)

\(\Rightarrow VT\ge\frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}\)

\(\Rightarrow VT\ge\frac{2\left(a+b+c+d\right)}{a+b+c+d}\)

\(\Rightarrow VT\ge2\)

\(\Rightarrow\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge2\)

\(\Leftrightarrow\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\ge2\) ( đpcm )

Lời giải:

Áp dụng bất đẳng thức AM-GM:

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

\(\sqrt{\frac{a}{b+c+d}}\geq \frac{2a}{a+b+c+d}\). Tương tự với các phân thức còn lại:

\(\Rightarrow \text{VT}\geq \frac{2(a+b+c+d)}{a+b+c+d}=2\) (đpcm)

tự diễn biến tự chuyển hóa là ko tốt :)

MK viết nhầm tất cả bỏ căn nhá

sai đề bài rồi trắc bạn viết nhầm

Đúng vậy, bỏ căn hết đi

Thôi làm luôn tìm ko ra thì chỉ phí time

Ta cm bổ đề  

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

\(=\frac{\frac{b+c+d}{a}+1}{2}\ge\sqrt{\frac{b+c+d}{a}}\) (đúng)

Tương tự cho 3 BĐT còn lại rồi cộng theo vế

\(VT\ge\frac{2a+2b+2c+2d}{a+b+c+d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2=VP\)

bài này vừa làm hôm qua xong chả nhớ ở web nào cả 

nhưng c/m \(\sqrt{\frac{a}{b+c+d}}\ge\frac{2a}{a+b+c+d}\) bằng AM-GM nhé đợi t tìm link cho đỡ phải làm lại

Dấu = xảy ra khi nào b???

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

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

Cộng vế với vế: \(VT\ge\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

Dấu "=" không xảy ra nên \(VT>2\)

a) Áp dụng bdt cosi schwars ta có 

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

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

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

bh mk can mn ho tro jup mk 2 cau cuoi nha

a) Áp dụng bđt Bunhiacopxki ta có :

\(\left[\left(\frac{a}{\sqrt{a+b}}^2\right)+\left(\frac{b}{\sqrt{b+c}}\right)^2+\left(\frac{c}{\sqrt{c+d}}\right)^2+\left(\frac{d}{\sqrt{d+a}}\right)\right]\)\(\times\)\(\left[\left(\sqrt{a+b}\right)^2+\left(\sqrt{b+c}\right)^2+\left(\sqrt{c+d}\right)^2+\left(\sqrt{d+a}\right)^2\right]\)\(\ge\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\right)\times2\left(a+b+c+d\right)\ge\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{a+b+c+d}{2}\)( chia cả 2 vế cho  \(2\left(a+b+c+d\right)\))

Dấu "=" xảy ra khi : a = b = c = d 

Vậy ...

Đặt \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=k\)\(\left(k>0\right)\)\(\Rightarrow\)\(a=Ak;b=Bk;c=Ck;d=Dk\)

\(\Rightarrow\)\(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=A\sqrt{k}+B\sqrt{k}+C\sqrt{k}+D\sqrt{k}\)

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

\(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{\left(Ak+Bk+Ck+Dk\right)\left(A+B+C+D\right)}\)

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

=> đpcm