\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)

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

Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)

=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)

\(\frac{\left(a+b\right)\left(1-c\right)}{2\left(a+b\right)\left(1-c\right)}+\frac{\left(b+c\right)\left(1-a\right)}{2\left(b+c\right)\left(1-a\right)}+\frac{\left(c+a\right)\left(1-b\right)}{2\left(c+a\right)\left(1-b\right)}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)

Ta chứng minh:\(\sqrt{a+bc}\ge a+\sqrt{bc}\)

\(\Leftrightarrow a+bc\ge a^2+bc+2a\sqrt{bc}\)

\(\Leftrightarrow a\ge a^2+2a\sqrt{bc}\)\(\Leftrightarrow a\ge a\left(a+2\sqrt{bc}\right)\Leftrightarrow1\ge a+2\sqrt{bc}\Leftrightarrow a+b+c\ge a+2\sqrt{bc}\)

\(\Leftrightarrow b+c-2\sqrt{bc}\ge0\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)(luôn đúng)

\(\Leftrightarrow\sqrt{a+bc}\ge a+\sqrt{bc}\)

CMTT\(\sqrt{b+ca}\ge b+\sqrt{ca}\)

          \(\sqrt{c+ab}\ge c+\sqrt{ab}\)

\(\Leftrightarrow\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)Vậy ......

(Dấu = xảy ra (=) a=b=c=1/3

Xét vế trái, ta có:  \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)(Do theo giả thiết thì ab + bc + bc = 1)

\(=\left(\frac{c}{a}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{a}{c}\right)+\left(\frac{b}{a}+\frac{b}{c}\right)+3\)

Khi đó, ta quy BĐT cần chứng minh về: \(\left(\frac{c}{a}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{a}{c}\right)+\left(\frac{b}{a}+\frac{b}{c}\right)\)\(\ge\sqrt{\frac{1}{a^2}+1}+\sqrt{\frac{1}{b^2}+1}+\sqrt{\frac{1}{c^2}+1}\)\(=\frac{\sqrt{a^2+1}}{a}+\frac{\sqrt{b^2+1}}{b}+\frac{\sqrt{c^2+1}}{c}\)

Theo BĐT Cauchy cho 2 số dương, ta có:

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

Tương tự ta có: \(\frac{\sqrt{b^2+1}}{b}\le\frac{2b+c+a}{2b}\)(2); \(\frac{\sqrt{c^2+1}}{c}\le\frac{2c+a+b}{2c}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được:

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

\(=3+\frac{1}{2}\left[\left(\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}\right)\right]\)

Đến đây, ta cần chứng minh \(\left(\frac{c}{a}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{a}{c}\right)+\left(\frac{b}{a}+\frac{b}{c}\right)\)\(\ge3+\frac{1}{2}\left[\left(\frac{c}{a}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{b}{c}\right)+\left(\frac{c}{a}+\frac{c}{b}\right)\right]\)

\(\Leftrightarrow\frac{1}{2}\left[\left(\frac{c}{a}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{b}{c}\right)+\left(\frac{c}{a}+\frac{c}{b}\right)\right]\ge3\)(Điều này hiển nhiên đúng vì theo BĐT Cauchy, ta có:

\(\frac{1}{2}\left[\left(\frac{c}{a}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{b}{c}\right)+\left(\frac{c}{a}+\frac{c}{b}\right)\right]\)\(\ge\frac{1}{2}.6\sqrt[6]{\frac{a^2b^2c^2}{a^2b^2c^2}}=3\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi a = b = c = \(\frac{1}{\sqrt{3}}\)

\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)

Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.

Lời giải vắn tắt:

\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)

( thay \(a^2+b^2+c^2=1\))

Bài này dùng AM-GM chắc cũng nhàm rồi nên em đổi kiểu nha.

\(VP-VT=\Sigma_{cyc}\frac{\left(ab+ac-2bc\right)^2+bc\left(b-c\right)^2}{2abc\left(b+c\right)\left(a^2+bc\right)}\ge0\)

<=>2ab+2bc+2ca<=1=1^2=(a+b+c)^2

<=>a^2+b^2+c^2+2ab+2bc+2ca>=2ab+2bc+2ca

<=>a^2+b^2+c^2>=0

a,b,c khong dong thoi =0

=> dang thuc khong xay ra

=> ab+bc+ca<1/2=>dpcm

(a+b+c)=1

a^2+b^2+c^2+2ab+2bc+2ca=1

a^^2+b^2+c^2>=0

=>2ab+2bc+2ca<=1

Đẳng thức khi (a+b+c=1 &0=> vô nghiệm

=> 2ab+2bc+2ca<1

=>ab+2bc+2ca<1/2

=>đpcm