\(3=a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\)\(abc\le1\)

\(VT=\frac{a^3\left(a+1\right)+b^3\left(b+1\right)+c^3\left(c+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a^4+b^4+c^4+a^3+b^3+c^3}{a+b+c+ab+bc+ca+abc+1}\)

\(\ge\frac{\frac{\left(a^2+b^2+c^2\right)^2}{3}+\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}}{\frac{\left(a+b+c\right)^2}{3}+5}=\frac{\frac{\frac{\left(a+b+c\right)^4}{9}}{3}+\frac{\frac{\left(a+b+c\right)^4}{9}}{3}}{8}\)

\(=\frac{\frac{\frac{3^4}{9}}{3}}{4}=\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

đề viết gì thế bạn ?

Ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\left(1\right)\)

Áp dụng bất đẳng thức buniacoxki ta có :

\(\left(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\right)\left(ab+bc+ac\right)\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)^2\)

Kết hợp với (1)

=> \(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

Nghe mùi holder ?

Áp dụng Svac + Cô-si 3 số được

\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=VP\left(đpcm\right)\)

"=" tại a = b = c

Áp dụng Svac

\(\Sigma\frac{a^3}{b+c}=\Sigma\frac{a^4}{ab+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\left(a^2+b^2+c^2\right)\)

"=" tại a=b=c

E thử làm cách khác ạ:))

Không mất tính tổng quát,giả sử \(a\ge b\ge c\)

\(\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}\ge\frac{b}{a+c}\ge\frac{c}{a+b}\end{cases}}\)

Áp dụng BĐT Trebysev ta có:

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

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

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

Dấu "=" xảy ra khi  \(a=b=c\)

\(=\frac{\left(\sin a+\cos a-\sin a+\cos a\right)\left(\sin a+\cos a+\sin a-\cos a\right)}{\sin a.\cos a}=\frac{2.\cos a.2.\sin a}{\sin a.\cos a}=4\)

\(\frac{cos\alpha}{1-\sin\alpha}=\frac{1+sin\alpha}{cos\alpha}\)

\(\Leftrightarrow cos^2\alpha=\left(1-sin\alpha\right)\left(1+\sin\alpha\right)\)

\(\Leftrightarrow cos^2\alpha=1-sin^2\alpha\)

\(\Leftrightarrow cos^2\alpha+sin^2\alpha=1\left(Đúng\right)\)