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

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

Với mọi số thực dương a;b;c ta có BĐT:

\(a^4+b^4\ge ab\left(a^2+b^2\right)\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Tương tự, ta có:

\(VT\le\dfrac{ab}{ab\left(a^2+b^2\right)+ab}+\dfrac{bc}{bc\left(b^2+c^2\right)+bc}+\dfrac{ca}{ca\left(c^2+a^2\right)+ca}\)

\(VT\le\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)

Đặt \(\left(a^2;b^2;c^2\right)=\left(x^3;y^3;z^3\right)\Rightarrow xyz=1\)

\(VT\le\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\)

Ta lại có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{xyz}{xy\left(x+y\right)+xyz}+\dfrac{xyz}{yz\left(y+z\right)+xyz}+\dfrac{xyz}{zx\left(z+x\right)+xyz}=1\)

Bài 1: 

Ta có: \(a+b\ge2\sqrt{ab}\)

\(b+c\ge2\sqrt{bc}\)

\(a+c\ge2\sqrt{ac}\)

Do đó: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

hay \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)

Đặt\(P=\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2+}+\dfrac{1}{2}\left(ab+bc+ca\right)\) 

Bổ đề: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\) \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) (1)

Chứng minh bổ đề: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\sqrt[3]{abc.\dfrac{1}{abc}}=9\left(\forall a,b,c\ge0\right)\) 

Kết hợp điều kiện đề bài ta được: \(a+b+c\ge3\)

Ta có: \(\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2\sqrt{b^2}}=\dfrac{ab}{2}\) ( AM-GM cho 2 số không âm 1 và b^2 )

\(\Rightarrow\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab}{2}\left(1\right)\)

Chứng minh hoàn toàn tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2}\left(2\right)\)

\(\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\left(3\right)\)

Cộng (1),(2),(3) vế theo vế thu được: \(P\ge a+b+c=3\)

Dấu "=" xảy ra tại a=b=c=1

Cách giải của

Ta có VP: 

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

Thay \(1=ab+bc+ca\)

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

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

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

\(=\dfrac{2}{\sqrt{\left[\left(a+c\right)\left(a+b\right)\left(b+c\right)\right]^2}}\)

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

_____________

Ta có VT: 

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

Thay \(1=ab+ac+bc\)

\(=\dfrac{a}{ab+ac+bc+a^2}+\dfrac{b}{ab+ac+bc+b^2}+\dfrac{c}{ab+ac+bc+c^2}\)

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

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

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

\(=\dfrac{ab+ac+ab+bc+ac+bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{2ab+2ac+2bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{2\cdot\left(ab+ac+bc\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\left(ab+ac+bc=1\right)\)

Mà: \(VP=VT=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

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

\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}\ge1\)

\(\Leftrightarrow2\ge\dfrac{a+b}{a+b+1}+\dfrac{b+c}{b+c+1}+\dfrac{c+a}{c+a+1}=\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+a+b}+\dfrac{\left(b+c\right)^2}{\left(b+c\right)^2+b+c}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)^2+c+a}\)

\(\Rightarrow2\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca+a+b+c}\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)+2\left(a+b+c\right)\ge2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)

\(\Rightarrow\)đpcm

Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)

Giả sử \(a\ge b\ge c\)thì

\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh

\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))

Vậy BĐT được chứng minh.

Dấu = xảy ra khi a=b=c