Vào đây nhé: https://hoc24.vn/hoi-dap/question/821240.html?pos=2125078

Mình đã trả lời rồi :3

Đặt: \(\frac{\left(a+b+c\right)^2}{ab+bc+ac}=t\)

Dễ chứng minh \(t\ge3\)

Ta viết lại biểu thức: \(\frac{\left(a+b+c\right)^2}{ab+bc+ac}+\frac{ab+bc+ac}{\left(a+b+c\right)^2}=t+\frac{1}{t}\)

\(=\frac{1}{9}t+\frac{1}{t}+\frac{8}{9}t\ge2\sqrt{\frac{1}{9}}+\frac{8}{9}t\ge\frac{2}{3}+\frac{24}{9}=\frac{10}{3}\)

\("="\Leftrightarrow t=3\Leftrightarrow a=b=c\)

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

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

Tương tự và cộng lại:

\(P+\frac{1}{2}\left(a+b+c\right)\ge\frac{3}{4}\left(a+b+c\right)\Rightarrow P\ge\frac{1}{4}\left(a+b+c\right)\)

\(\Rightarrow P\ge\frac{1}{4}\sqrt{3\left(ab+bc+ca\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(gt\Rightarrow\frac{ab+bc+ca}{abc}=0\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-czx\right)+3xyz\)

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

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

\(=abc\left[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\right]\)

\(=abc\cdot\frac{3}{abc}=3\)

Ta có:

ab + ac + bc = 0

\(\Rightarrow\) \(\frac{ab+ac+bc}{abc}=0\)

\(\Rightarrow\) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Đặt \(\frac{1}{a}=x;\) \(\frac{1}{b}=y;\) \(\frac{1}{c}=z\)

Mà x + y + z = 0

=> x³ + y³ + z³ = 3xyz (Tự chứng minh nhé bạn, nếu không chứng minh được thì bình luận nhé!)

\(\Rightarrow\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Ta có:

\(A=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}\)

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

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

\(A=abc.\frac{3}{abc}\)

\(A=3\)

Tham khảo: Câu hỏi của Đậu Đình Kiên

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham

ta có \(\frac{11b^3-a^3}{ab+4b^2}+\frac{11c^3-b^3}{bc+4c^2}+\frac{11a^3-c^3}{ca+4a^2}=\frac{11-\left(\frac{a}{b}\right)^3}{\frac{a}{b}+4}\cdot b+\frac{11-\left(\frac{b}{c}\right)^3}{\frac{b}{c}+4}\cdot c+\frac{11-\left(\frac{c}{a}\right)^3}{\frac{c}{a}+4}\cdot a\)

khi a=b=c=1 ta thấy đẳng thức xảy ra

xét \(f\left(x\right)=\frac{11-x^3}{x+4}\)ta có \(\frac{11-x^3}{x+4}\le-x+3\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\ge0\forall x>0\)

thay x bởi a/_b ta được \(\frac{11-\left(\frac{a}{b}\right)^3}{\frac{a}{b}+4}\le-\frac{a}{b}+3\Leftrightarrow\frac{11b^3-a^3}{ab+4b^2}\le-a+3b\)

tương tự \(\hept{\begin{cases}\frac{11c^3-b^3}{bc+4c^2}\le-b+3c\\\frac{11ba^3-c^3}{ac+4a^2}\le-c+3a\end{cases}}\)

cộng các bđt cùng chiều ta được

\(\frac{11b^3-a^3}{ab+4b^2}+\frac{11c^3-b^3}{bc+4c^2}+\frac{11a^3-c^3}{ac+4a^2}\le2\left(a+b+c\right)=6\)

\(\frac{11b^3-a^3}{ab+4b^2}\le3b-a\)

Câu hỏi của TRẦN HỮU ĐẠT - Toán lớp 9 - Học toán với OnlineMath