chết đăng nhầm sogy nha

\(\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}+\dfrac{\left(c+a\right)^2}{5b^2+\left(c+a\right)^2}+\dfrac{\left(a+b\right)^2}{5c^2+\left(a+b\right)}\ge\dfrac{4}{3}\)

\(\Leftrightarrow\dfrac{-20a^2+10bc+5b^2+c^2}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{-20b^2+10ac+5c^2+5a^2}{9\left(5b^2+\left(c+a\right)^2\right)}+\dfrac{-20c^2+10ab+5a^2+5b^2}{9\left(5c^2+\left(a+b\right)\right)}\ge0\)

\(\Leftrightarrow\sum_{cyc}\dfrac{\left(c-a\right)\left(10a+5b+5c\right)-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\dfrac{-\left(a-b\right)\left(10a+5b+5c\right)}{9\left(5a^2+\left(b+c\right)^2\right)}+\dfrac{\left(a-b\right)\left(10b+5a+5c\right)}{9\left(5b^2+\left(a+c\right)^2\right)}\right)\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)\left(\dfrac{10b+5a+5c}{9\left(5b^2+\left(a+c\right)^2\right)}-\dfrac{10a+5b+5c}{9\left(5a^2+\left(b+c\right)^2\right)}\right)\right)\ge0\)

\(\Leftrightarrow\sum_{cyc}\left(\left(a-b\right)^2\dfrac{5\left(a^2+b^2-c^2+4ab\right)}{3\left(a^2+2ac+5b^2+c^2\right)\left(5a^2+b^2+2bc+c^2\right)}\right)\ge0\)

Dau "=" khi \(a=b=c\)

Bất đẳng thức cần chứng minh tương đương với

\(\dfrac{4}{3}-\sum\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}\le0\Leftrightarrow1-\sum\dfrac{\left(b+c\right)^2}{5a^2+\left(b+c\right)^2}\le\dfrac{5}{3}\Leftrightarrow\sum\dfrac{5a^2}{5a^2+\left(b+c\right)^2}\le\dfrac{1}{3}\)\(\dfrac{9}{5a^2+\left(b+c\right)^2}=\dfrac{\left(1+2\right)^2}{a^2+b^2+c^2+2\left(2a^2+bc\right)}\le\dfrac{1}{a^2+b^2+c^2}+\dfrac{2}{2a^2+bc}\)

\(\Rightarrow\sum\dfrac{9a^2}{5a^2+\left(b+c\right)^2}\le\sum\dfrac{a^2}{a^2+b^2+c^2}+\sum\dfrac{2a^2}{2a^2+bc}=4-\sum\dfrac{bc}{2a^2+bc}\)Cần chứng minh \(\sum\dfrac{bc}{2a^2+bc}\ge1\). Ta có:

\(\sum\dfrac{bc}{2a^2+bc}\ge\dfrac{\left(\sum bc\right)^2}{\sum bc\left(2a^2+bc\right)}=1\)

Đẳng thức xảy ra khi \(a=b=c\) hoặc \(a=0;b=c\) và các hoán vị

Giải theo kiểu lớp 8 cho chắc :v

Ta có : \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)

\(\Leftrightarrow\dfrac{3a^2+3b^2+3c^2}{9}\ge\dfrac{\left(a+b+c\right)^2}{9}\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ( Đúng )

Vậy BĐT đã được chứng minh . Dấu \("="\) xảy ra khi \(a=b=c\)

Áp dụng BĐT Cauchy - schwarz dưới dạng engel ta có :

\(\dfrac{a^2+b^2+c^2}{3}=\dfrac{a^2}{3}+\dfrac{b^2}{3}+\dfrac{c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}=\left(\dfrac{a+b+c}{3}\right)^2\)

Dấu \("="\) xảy ra khi \(a=b=c\)

Bài này đã có ở đây:

Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24

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)