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

=> 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 

Không làm mất tính tổng quát của bài toán, giả sử \(a\ge b\ge c\)(1)

Có \(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}=\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)

Từ (1) => \(\hept{\begin{cases}\frac{2}{a}\le\frac{1}{a}+\frac{1}{b}\\\frac{2}{b}\le\frac{1}{b}+\frac{1}{c}\\\frac{2}{c}\le\frac{1}{a}+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{\frac{2}{a}}\le\sqrt{\frac{1}{a}+\frac{1}{b}}\\\sqrt{\frac{2}{b}}\le\sqrt{\frac{1}{b}+\frac{1}{c}}\\\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{a}+\frac{1}{c}}\end{cases}}\)

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

=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)

Ta có đpcm

3a) ta có \(\frac{a^2}{a+b}=a-\frac{ab}{a+b}>=a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)

vì \(a,b>0,a+b>=2\sqrt{ab}nên\frac{ab}{a+b}< =\frac{ab}{2\sqrt{ab}}\)

tương tự \(\frac{b^2}{b+c}=b-\frac{bc}{b+c}>=b-\frac{bc}{2\sqrt{bc}}=b-\frac{\sqrt{bc}}{2}\)

tương tự \(\frac{c^2}{c+a}=c-\frac{ca}{c+a}>=c-\frac{ca}{2\sqrt{ca}}=c-\frac{\sqrt{ca}}{2}\)

cộng từng vế BĐT ta được \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=a+b+c-\frac{\sqrt{ab}}{2}-\frac{\sqrt{bc}}{2}-\frac{\sqrt{ca}}{2}=\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}\left(1\right)\)

giả sử \(\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}>=\frac{a+b+c}{2}\)

<=> \(2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=a+b+c\)

<=> \(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=0\)

<=> \(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}>=0\)

<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)

(đúng với mọi a,b,c >0) (2)

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

Ta có: \(\frac{a^3+b^3}{\sqrt{a^2-ab+b^2}}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\sqrt{a^2-ab+b^2}}=\left(a+b\right)\sqrt{a^2-ab+b^2}\)

\(=\sqrt{a+b}\sqrt{\left(a+b\right)\left(a^2-ab+b^2\right)}=\sqrt{a+b}\sqrt{a^3+b^3}\)

\(=\sqrt{\left(a+b\right)\left(a^3+b^3\right)}=\sqrt{\left(\sqrt{a}^2+\sqrt{b}^2\right)\left(\sqrt{a^3}^{^2}+\sqrt{b^3}^{^2}\right)}\)

Áp dụng BĐT Bunhi... ta có:

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

\(\Rightarrow\sqrt{\left(\sqrt{a}^2+\sqrt{b}^2\right)+\left(\sqrt{a^3}^{^2}+\sqrt{b^3}^{^2}\right)}\)\(\ge\sqrt{a}\sqrt{a^3}+\sqrt{b}\sqrt{b^3}=\sqrt{a^4}+\sqrt{b^4}=a^2+b^2\)

\(\Rightarrow\frac{a^3+b^3}{\sqrt{a^2-ab+b^2}}\ge a^2+b^2\) (1)

Tương tự ta có: \(\frac{b^3+c^3}{\sqrt{b^2-bc+c^2}}\ge b^2+c^2\) (2)

\(\frac{c^3+d^3}{\sqrt{c^2-cd+d^2}}\ge c^2+d^2\)(3)

\(\frac{d^3+a^3}{\sqrt{d^2-da+a^2}}\ge d^2+a^2\)(4)

Cộng vế với vế của 1,2,3,4 ta được:

\(\frac{a^3+b^3}{\sqrt{a^2-ab+b^2}}+\frac{b^3+c^3}{\sqrt{b^2-bc+c^2}}+\frac{c^3+d^3}{\sqrt{c^2-cd+d^2}}+\frac{d^3+a^3}{\sqrt{d^2-da+a^2}}\)\(\ge2\left(a^2+b^2+c^2+d^2\right)\left(\text{đ}pcm\right)\)

Hoặc \(\left(a+b\right)\sqrt{a^2-ab+b^2}\ge a^2+b^2\Leftrightarrow ab\left(a-b\right)^2\ge0\)(bình phương lên)

Áp dụng bđt AM-GM:

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

Tương tự và cộng theo vế:

\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu "=" không xảy ra => đpcm

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

Tương tự \(\sqrt{\frac{b}{1-b}}\ge\frac{2b}{a+b+c}\) và \(\sqrt{\frac{c}{1-c}}\ge\frac{2c}{a+b+c}\)

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

Dấu "=" xảy ra \(\Leftrightarrow a=b+c;b=a+c;c=a+b\Rightarrow a+b+c=0\) (KTM)

Vậy \(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}>2\)

BĐT Nesbitt cho 4 biến, bạn tham khảo google nhiều lắm :3

Mk viết nhầm tất cả bỏ căn nhá

đề bài sai rồi bạn nhé check lại đi 

Sửa đề: \(\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}\ge\sqrt{2}\left(\Sigma\sqrt{\frac{1-a}{a}}\right)\)

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

Theo AM-GM:\(\frac{b+c}{a}\ge2\sqrt{\frac{2\left(b+c\right)}{a}}-2\)

Tương tự và cộng lại: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-6\)

Mà: \(\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\ge3\sqrt[6]{\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\ge6\)

Từ đó: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}=VP\)

Done!