Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$\Rightarrow ab+bc+ac=0$

Đặt $ab=x, bc=y, ac=z$ thì $x+y+z=0$

Có:

$M=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}$
$=\frac{b^3c^3+a^3c^3+a^3b^3}{(abc)^2}$

$=\frac{x^3+y^3+z^3}{xyz}=\frac{(x+y)^3-3xy(x+y)+z^3}{xyz}$

$=\frac{(-z)^3-3xy(-z)+z^3}{xyz}$
$+\frac{-z^3+3xyz+z^3}{xyz}=\frac{3xyz}{xyz}=3$

xí câu 1:))

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)(1)

Đặt a = x + y - 2 => a > 0 ( vì x,y > 1 )

Khi đó \(\left(1\right)=\frac{\left(a+2\right)^2}{a}=\frac{a^2+4a+4}{a}=\left(a+\frac{4}{a}\right)+4\ge2\sqrt{a\cdot\frac{4}{a}}+4=8\)( AM-GM )

Vậy ta có đpcm

Đẳng thức xảy ra <=> a=2 => x=y=2

Câu hỏi của Conan Kudo - Toán lớp 8 - Học toán với OnlineMath

Bạn tham khảo nhé!

\(a\text{) }\)Áp dụng: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (a, b > 0). Dấu "=" xảy ra khi a = b.

\(\frac{1}{a^2+b^2}+\frac{1}{ab}=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2.\frac{\left(a+b\right)^2}{4}}=\frac{6}{\left(a+b\right)^2}\)

\(=6\left[\frac{1}{\left(a+b\right)^2}+\frac{27}{8}\left(a+b\right)+\frac{27}{8}\left(a+b\right)\right]-\frac{81}{2}\left(a+b\right)\)

\(\ge6.3\sqrt[3]{\frac{1}{\left(a+b\right)^2}.\frac{27}{8}\left(a+b\right).\frac{27}{8}\left(a+b\right)}-\frac{81}{2}\left(a+b\right)\)

\(=\frac{81}{2}-\frac{81}{2}\left(a+b\right)\)

Tương tự: \(\frac{1}{b^2+c^2}+\frac{1}{bc}\ge\frac{81}{2}-\frac{81}{2}\left(b+c\right)\)

\(\frac{1}{c^2+a^2}+\frac{1}{ca}\ge\frac{81}{2}-\frac{81}{2}\left(c+a\right)\)

Cộng theo vế ta được 

\(A\ge3.\frac{81}{2}-81\left(a+b+c\right)=3.\frac{81}{2}-81=\frac{81}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}.\)

Vậy GTNN của A là \(\frac{81}{2}.\)

Ta có: \(0\le a\le b\le1.\)

\(\Rightarrow\left\{{}\begin{matrix}a-1\le0\\b-1\le0\end{matrix}\right.\)

\(\Rightarrow\left(a-1\right).\left(b-1\right)\ge0\)

\(\Rightarrow ab-a-b+1\ge0.\)

\(\Rightarrow ab+1\ge0+a+b\)

\(\Rightarrow ab+1\ge a+b\)

\(\Rightarrow\frac{1}{ab+1}\le\frac{1}{a+b}.\)

\(\Rightarrow\frac{c}{ab+1}\le\frac{c}{a+b}\left(c\ge0\right).\)

Mà \(\frac{c}{a+b}\le\frac{2c}{a+b+c}\left(c\ge0\right)\)

\(\Rightarrow\frac{c}{ab+1}\le\frac{2c}{a+b+c}\left(1\right).\)

Chứng minh tương tự ta cũng có:

\(\frac{b}{ac+1}\le\frac{2b}{a+b+c}\left(2\right);\frac{a}{bc+1}\le\frac{2a}{a+b+c}\left(3\right).\)

Cộng theo vế \(\left(1\right);\left(2\right)và\left(3\right)\) ta được:

\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2a+2b+2c}{a+b+c}\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2.\left(a+b+c\right)}{a+b+c}\)

\(\Rightarrow\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le2\left(đpcm\right).\)

Chúc bạn học tốt!