Bạn CM \(a^5+b^5\ge ab\left(a^3+b^3\right)\)

\(\Rightarrow\frac{ab}{a^5+b^5+ab}\le\frac{1}{a^3+b^3+abc}\)

Tiếp tục \(a^3+b^3\ge ab\left(a+b\right)\)

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

\(\Rightarrow\frac{ab}{a^5+b^5+ab}\le\frac{c}{a+b+c}\)

Tương tự cộng lại suy ra \(VT\le1\)

Dấu = xảy ra khi a=b=c=1

Mỉnh cảm ơn nha 

Ta có :

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

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

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

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

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

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

Ta lại có :

\(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{bca}{b^3}+\frac{cab}{c^3}\)

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

\(\)

Bài làm:

Ta có: \(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)

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

CM HĐT phụ:

Ta có: \(a^3+b^3+c^3=\left(a^3+b^3+c^3-3abc\right)+3abc\)

\(=\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\right]+3abc\)

\(=\left[\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\right]+3abc\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)

Áp dụng vào trên ta được:

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

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

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

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

Vậy P = 3

Từ \(a^5+b^5=\left(a+b\right)\left(a^4-a^3b+a^2b^2-ab^3+b^4\right)\)

\(=\left(a+b\right)\left[a^2b^2+a^3\left(a-b\right)-b^3\left(a-b\right)\right]\)

\(=\left(a+b\right)\left[a^2b^2+\left(a-b\right)\left(a^3-b^3\right)\right]\)

\(=\left(a+b\right)\left[a^2b^2+\left(a-b\right)^2\left(a^2+ab+b^2\right)\right]\)

\(\ge\left(a+b\right)^2a^2b^2\forall a,b>0\)

\(\Rightarrow a^5+b^5+ab\ge ab\left[ab\left(a+b\right)+1\right]\)

\(\Rightarrow\frac{1}{a^5+b^5+ab}\le\frac{1}{ab\left(a+b\right)+1}=\frac{c}{a+b+c}\left(abc=1\right)\)

Tương tự cũng có: \(\frac{bc}{b^5+c^5+bc}\le\frac{a}{a+b+c};\frac{ca}{c^5+a^5+ca}\le\frac{b}{a+b+c}\)

Cộng theo vế ta có: 

\(VT\le\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

Đẳng thức xảy ra khi \(a=b=c=1\)

ab+bc+ca=414

=>2a+2b+2c=414

=>2(a+b+c)=414

=>a+b+c=207

Áp dụng t/c của dãy tỉ số bằng nhau, ta có

a/2=b/3=c/8=a+b+c/2+3+8=207/13=15,9

a/2=15,9=>a=31,8

b/3=15,9=>b=47,7

c/8=15,9=>c=127,2

Kết luận

tk mk nha

Áp dụng BĐT AM-GM (Cô si): \(A\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(=3\sqrt[3]{\frac{1}{a\left(b+c\right).b\left(c+a\right).c\left(a+b\right)}}=\frac{3}{\sqrt[3]{\left(ab+ca\right)\left(bc+ab\right)\left(ca+bc\right)}}\)

\(\ge\frac{9}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

Đẳng thức xảy ra khi a = b = c = 1

P/s: Check giúp em xem có ngược dấu không:v

Cach khac 

Dat \(\left(ab;bc;ca\right)\rightarrow\left(x;y;z\right)\)

\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x^2+y^2+z^2\ge3\\xyz\le1\end{cases}}\)

Ta co:

\(A=\frac{1}{ab+b^2}+\frac{1}{bc+c^2}+\frac{1}{ca+a^2}\)

\(=\frac{1}{x+\frac{xy}{z}}+\frac{1}{y+\frac{yz}{x}}+\frac{1}{z+\frac{zx}{y}}\ge\frac{9}{3+xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\ge\frac{9}{3+3}=\frac{3}{2}\)

Dau '=' xay ra khi \(a=b=c=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(a=b=c=1\)

MaiLink bạn hãy chứng minh: \(xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le3\) xem thế nào? Nếu như ko c/m được thì bài này ngược dấu.