viết thế nay bố ai hiểu được

bạn kì quá ko giúp thì thôi còn phàn nàn. 

Bđt \(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)=\Sigma_{cyc}\left(\frac{a\left(a+b\right)\left(a-b\right)}{2\left(a^2+b^2\right)}\right)\)

\(=\Sigma_{cyc}\left(a-b\right)\left(\frac{a\left(a+b\right)}{2\left(a^2+b^2\right)}-\frac{1}{2}\right)+\frac{1}{2}\Sigma_{cyc}\left(a-b\right)\)

\(=\Sigma_{cyc}\frac{b\left(a-b\right)^2}{2\left(a^2+b^2\right)}\ge0\) (đúng)

Đẳng thức xảy ra khi a = b = c = d 

P/s: Em thử, sai thì thôi nha!

\(VT-VP=\Sigma_{cyc}\frac{2a+b+c}{a^2b\left(a+b+c\right)}\left(a-b\right)^2\ge0\)

hay \(\frac{a}{c^2}+\frac{1}{a}\ge\frac{2}{c}\)\(\Leftrightarrow\)\(\frac{a}{c^2}\ge\frac{2}{c}-\frac{1}{a}\)\(\Rightarrow\)\(VT\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

"=" \(\Leftrightarrow\)\(a=b=c\)

\(\Leftrightarrow\frac{a^2}{3}+b^2+4c^2-ab-2bc-2ca>0\)

\(\Leftrightarrow\frac{a^2}{4}+\left(b^2+4bc+4c^2\right)-a\left(b+2c\right)+\frac{a^2}{12}-6bc>0\)

\(\Leftrightarrow\frac{a^2}{4}+\left(b+2c\right)^2-a\left(b+2c\right)+\frac{a^2-36bc}{12}>0\)

\(\Leftrightarrow\left(\frac{a}{2}-b-2c\right)^2+\frac{a^3-36abc}{12a}>0\)

\(\Leftrightarrow\left(\frac{a}{2}-b-2c\right)^2+\frac{a^3-36}{12a}>0\) (1)

Do \(a^3>36\Rightarrow\left\{{}\begin{matrix}a>0\\a^3-36>0\end{matrix}\right.\) \(\Rightarrow\frac{a^3-36}{12a}>0\)

\(\Rightarrow\left(1\right)\) luôn đúng

Ta có BĐT \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

Nên BĐT cần chứng minh là 

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

Đặt \(\hept{\begin{cases}a^2=x\\b^2=y\\c^2=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x,y,z>0\end{cases}}\)

Áp dụng BĐT AM-GM and Cauchy-Schwarz ta có:

\(Σ\frac{a^2}{a+b^2}=Σ\frac{x}{\sqrt{x}+y}=Σ\frac{x}{\sqrt{\frac{x\left(x+y+z\right)}{3}+y}}\)

\(=Σ\frac{6x}{2\sqrt{3x\left(x+y+z\right)}+6y}\geΣ\frac{6x}{3x+x+y+z+6y}=Σ\frac{6x}{4x+7y+z}\)

\(=Σ\frac{6x^2}{4x^2+7xy+xz}\ge\frac{6\left(x+y+z\right)^2}{Σ\left(4x^2+7xy+xz\right)}=\frac{3}{2}\)

-Nguồn : Xem câu hỏi

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

\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)

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

Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrowđpcm\)

\(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge3\sqrt[3]{\frac{a^3}{b+c}.\frac{a^3}{b+c}.\frac{\left(b+c\right)^2}{8}}=\frac{3a^2}{2}\)

Rồi tương tự các kiểu:v

Suy ra \(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)

\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}=\left(a^2+b^2+c^2\right)\) (chú ý \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))

Không phải dùng tới Cauchy-Schwarz:D

mình chưa hiểu?

có thể giải thích rõ hơn đc ko