Bài dễ lắm làm đi hỏi làm gì

Lại gặp thánh troll rồi

\(\frac{a^3}{b^2+3}=\frac{a^3}{b^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)

Tương tự

\(\Rightarrow\Sigma_{cyc}\frac{a^3}{b^2+3}=\Sigma_{cyc}\frac{a^3}{\left(a+b\right)\left(b+c\right)}\)

Theo Cô-si:\(\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3}{4}a\)

\(\Rightarrow\Sigma_{cyc}\frac{a^3}{\left(a+b\right)\left(b+c\right)}\ge\frac{1}{4}\left(a+b+c\right)\ge\frac{1}{4}\sqrt{3\left(ab+bc+ca\right)}=\frac{3}{4}\)

\(\frac{1}{a^2+b^2+2}+\frac{1}{b^2+c^2+2}+\frac{1}{c^2+a^2+2}\le\frac{3}{4}\)

\(\Leftrightarrow\frac{a^2+b^2}{a^2+b^2+2}+\frac{b^2+c^2}{b^2+c^2+2}+\frac{c^2+a^2}{c^2+a^2+2}\ge\frac{3}{2}\)

Áp dụng BĐT Cauchy - Schwarz ta có :

\(VT\ge\frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\)

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

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

Ta cần chứng minh :

\(\frac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\ge\frac{3}{2}\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge0\) luôn đúng 

Chúc bạn học tốt !!!

hoang viet nhat copy nhớ ghi nguồn nha bạn:))Link 

Mà quan trọng là copy mà bạn có hiểu không là chuyện khác:) Bạn hãy giải thích tại sao:

\(\frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\ge\frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\right)}{a^2+^2+c^2}\)

Hm hình như bài này t làm 1 lần rồi mà sao nó ko hiển thị lên ở phần câu hỏi tương tự nhỉ ??

Ta có: \(a^2+1\ge2a\)

\(\Leftrightarrow a^2+2b+3\ge2\left(a+b+1\right)\)

\(\Leftrightarrow\frac{a}{a^2+2b+3}\le\frac{a}{2\left(a+b+1\right)}\)

Tương tự rồồi cộng theo vế ta được: 

\(VT\le\frac{1}{2}\left(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\right)\)

Ta cần chứng minh \(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)

\(\Leftrightarrow-\left(\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\right)\le-2\)

\(\Leftrightarrow\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\ge2\)

Áp dụng BĐT Cauchy-Schwarz ta có:

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

\(\ge\frac{\left(a+b+c+3\right)^2}{a^2+b^2+c^2+3\left(a+b+c\right)+ab+bc+ca+3}\)

\(=\frac{\left(a+b+c+3\right)^2}{\frac{1}{2}\left(a^2+b^2+c^2+6\left(a+b+c\right)+2\left(ab+bc+ca\right)+9\right)}\)

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

Dấu "=" <=> a=b=c=1

\(\frac{a}{b^2+1}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}\ge a-\frac{ab}{2}\)  (AM-GM)

chung minh tuong tu ta co 

\(VT\ge a+b+c-\frac{ab}{2}-\frac{bc}{2}-\frac{ac}{2}\ge3-\frac{\left(a+b+c\right)^2}{6}\ge3-\frac{3}{2}=\frac{3}{2}\)

dau = xay ra khi a=b=c=1

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

<=>  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab+bc+ca\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

<=>  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab+bc+ca\right)\ge9\)

Ap dung BDT AM-GM ta co:

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

\(=\frac{3}{abc}+\left(ab+bc+ca\right)+\left(ab+bc+ca\right)\)

\(\ge3\sqrt[3]{\frac{3}{abc}\left(ab+bc+ca\right)\left(ab+bc+ca\right)}\)

\(\ge3\sqrt[3]{\frac{3}{abc}.3abc\left(a+b+c\right)}=9\)

=>  dpcm