Bài này có bạn giải rồi:

Cho các số thực dương a,b,c.Chứng minh rằng :\(\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}+\dfrac{c\left(2b-c\right)}{... - Hoc24

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)

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

Nên ta chỉ cần chứng minh:

\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)

Điều này hiển nhiên đúng do:

\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)

Dấu "=" xảy ra khi \(a=b=c\)

Trước hết theo BĐT Schur bậc 3 ta có:

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)+9abc\ge2\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+3abc\ge2\left(ab+bc+ca\right)\) (do \(a+b+c=3\)) (1)

Đặt vế trái BĐT cần chứng minh là P, ta có:

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

\(\Rightarrow P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)

Áp dụng (1):

\(\Rightarrow P\ge\dfrac{\left[2\left(ab+bc+ca\right)\right]^2}{\left(ab+bc+ca\right)^2}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đề bài sai với \(a=b=c=2\)

đề đúng nhớ áp dụng AM-GM

AD bđt AM-GM cho 3 số

\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)

\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)

thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}-2+\dfrac{c\left(2b-c\right)}{b\left(c+a\right)}-2+\dfrac{a\left(2c-a\right)}{c\left(a+b\right)}-2\le\dfrac{3}{2}-6\)

\(\Leftrightarrow\dfrac{b^2+2ac}{a\left(b+c\right)}+\dfrac{c^2+2ab}{b\left(c+a\right)}+\dfrac{a^2+2bc}{c\left(a+b\right)}\ge\dfrac{9}{2}\)

\(\Leftrightarrow\dfrac{b^2}{ab+ac}+\dfrac{c^2}{bc+ab}+\dfrac{a^2}{ac+bc}+\dfrac{2c^2}{bc+c^2}+\dfrac{2a^2}{ac+a^2}+\dfrac{2b^2}{ab+b^2}\ge\dfrac{9}{2}\)

Ta có:

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

\(\Leftrightarrow VT\ge\left(a+b+c\right)^2\left(\dfrac{1}{2\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2+ab+bc+ca}\right)\)

\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+2\left(a^2+b^2+c^2+ab+bc+ca\right)}\)

\(\Leftrightarrow VT\ge\dfrac{9\left(a+b+c\right)^2}{2\left(a+b+c\right)^2}=\dfrac{9}{2}\)

đăng câu khác đi câu này nổi tiếng rồi

APMO 2005

Áp dụng BDT AM-GM:

\(\sqrt{a^3+1}=\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\le\dfrac{1}{2}\left(a^2+2\right)\)

tương tự như thế, ta có:

\(VT=\sum\dfrac{a^2}{\sqrt{\left(a^3+1\right)\left(b^3+1\right)}}\ge\sum\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\)

Cần chứng minh \(\sum\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\ge\dfrac{4}{3}\Leftrightarrow\sum\dfrac{a^2}{\left(a^2+2\right)\left(b^2+2\right)}\ge\dfrac{1}{3}\)

\(\Leftrightarrow3\left[a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)\right]\ge\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\)

\(\Leftrightarrow3\left[a^2b^2+b^2c^2+c^2a^2+2\left(a^2+b^2+c^2\right)\right]\ge a^2b^2c^2+2\left(a^2b^2+b^2c^2+c^2a^2\right)+4\left(a^2+b^2+c^2\right)+8\)

\(\Leftrightarrow\left(a^2b^2+b^2c^2+c^2a^2\right)+2\left(a^2+b^2+c^2\right)\ge72\)

Điều này đúng theo AM-GM:

\(\left\{{}\begin{matrix}a^2b^2+b^2c^2+c^2a^2\ge3\sqrt[3]{a^4b^4c^4}=48\\a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}=12\end{matrix}\right.\)(abc=8)

Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=2

P/s : góp thêm đề :

1) \(\sum\sqrt{\dfrac{a}{b+3c}}\ge\dfrac{3}{2}\)

2)\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{2\left(ab+bc+ca\right)}{a^2+b^2+c^2}\ge5\)

3) \(a^3+b^3+c^3=3.\)CMR:\(\sum\dfrac{a}{b+2}\le1\)

4) CMr :\(\sum\dfrac{a^2}{a^2+b^2}\ge\sum\dfrac{a}{a+b}\)

Chuẩn hóa \(a+b+c=3\)

\(\dfrac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}=\dfrac{\left(a+3\right)^2}{2a^2+\left(3-a\right)^2}=\dfrac{a^2+6a+9}{3\left(a^2-2a+3\right)}=\dfrac{1}{3}\left(1+\dfrac{8a+6}{\left(a-1\right)^2+2}\right)\le\dfrac{1}{3}\left(1+\dfrac{8a+6}{2}\right)\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3}\left(3+\dfrac{8\left(a+b+c\right)+18}{2}\right)=8\) (đpcm)