Áp dụng BĐT Cauchy-Schwarz:

$\frac{a}{a+\sqrt{2016a + bc}}=\frac{a}{a+\sqrt{(a+b+c)a + bc}} =\frac{a}{a+\sqrt{(a+b)(c+a)}} \leq \frac{a}{a+\sqrt{(\sqrt{ab}+\sqrt{ac})^{2}}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$

$\Rightarrow \frac{a}{a+\sqrt{2016a + bc}} + \frac{b}{b+\sqrt{2016b + ca}} + \frac{c}{c+\sqrt{2016c + ab}}\leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$

...............................

giúp với m.n

Làm đơn giản thế này thôi nhé An Kì :

Ta có : \(2016a+bc=\left(a+b+c\right)a+bc=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)Tương tự : \(2016b+ac=\left(a+b\right)\left(b+c\right)\)

\(2016c+ab=\left(a+c\right)\left(b+c\right)\)

\(\Rightarrow\left(2016a+bc\right)\left(2016b+ac\right)\left(2016c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

áp dụng tính chất dảy tỉ số bằng nhau ta có :

\(\dfrac{2016c-a-b+2016b-a-c+2016a-b-c}{c+b+a}\)

\(=\dfrac{2014c+2014b+2014a}{c+b+a}=\dfrac{2014\left(c+a+b\right)}{c+a+b}=2014\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2016c-a-b}{c}=2014\\\dfrac{2016b-a-c}{b}=2014\\\dfrac{2016a-b-c}{a}=2014\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2016c-a-b=2014c\\2016b-a-c=2014b\\2016a-b-c=2014a\end{matrix}\right.\)

bấm máy tính ta có phương trình vô nghiệm nên A không xát định

\(L=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

Khi đó \(L=\dfrac{-abc}{abc}=-1\)

Với \(a+b+c\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{2016c-a-b}{c}=\dfrac{2016b-a-c}{b}=\dfrac{2016a-b-c}{a}=\dfrac{2016c-a-b+2016b-a-c+2016c-b-c}{a+b+c}=\dfrac{\left(2016c-c-c\right)+\left(2016b-b-b\right)+\left(2016c-c-c\right)}{a+b+c}=\dfrac{2014\left(a+b+c\right)}{a+b+c}=2014\)\(\Rightarrow\left\{{}\begin{matrix}2016c-a-b=2014c\\2016b-a-c=2014b\\2016a-b-c=2014a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2016c-a-b-2014c=0\\2016b-a-c=2014b=0\\2016a-b-c-2014a=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2c-a-b=0\\2b-a-c=0\\2a-b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)

Khi đó \(L=\dfrac{8abc}{abc}=8\)

\(A=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

\(A=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

\(A=\dfrac{a^2bc}{ab\left(1+ac+c\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)

\(A=\dfrac{ac+1+c}{ac+c+1}\)

\(A=1\)

\(A=\dfrac{ab}{ab+a+1}+\dfrac{bc}{bc+b+1}+\dfrac{ca}{ca+c+1}\)

\(A=\dfrac{abc}{abc+ac+c}+\dfrac{bc}{bc+b+abc}+\dfrac{ca}{ca+c+1}\)

\(A=\dfrac{1}{1+ac+c}+\dfrac{c}{c+1+ac}+\dfrac{ca}{ca+c+1}\)

\(A=1\)

Lười đánh máy thật sự, buốt tay lắm:((

Ta có: \(Q=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

\(Q=\dfrac{ac}{c\left(ab+a+1\right)}+\dfrac{abc}{ac\left(bc+b+1\right)}+\dfrac{c}{ac+c+1}\)

\(Q=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}\)

\(Q=\dfrac{ac}{1+ac+c}+\dfrac{1}{c+a+ac}+\dfrac{c}{ac+c+1}\)

\(Q=\dfrac{ac+1+c}{1+ac+c}=1\)

Vậy Q=1

Q=ab+a+1a​+bc+b+1b​+ac+c+1c​

Q=\dfrac{ac}{c\left(ab+a+1\right)}+\dfrac{abc}{ac\left(bc+b+1\right)}+\dfrac{c}{ac+c+1}Q=c(ab+a+1)ac​+ac(bc+b+1)abc​+ac+c+1c​

Q=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}Q=abc+ac+cac​+abc2+abc+acabc​+ac+c+1c​

Q=\dfrac{ac}{1+ac+c}+\dfrac{1}{c+a+ac}+\dfrac{c}{ac+c+1}Q=1+ac+cac​+c+a+ac1​+ac+c+1c​

Q=\dfrac{ac+1+c}{1+ac+c}=1Q=1+ac+cac+1+c​=1

chúc bạn thi tốt

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)

Lời giải:

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

\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)

\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)

\(\Rightarrow a=b=c\) (do $a,b,c>0$)

$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$

tham khảo bài tương tự này :  

Câu hỏi của so yeoung cheing - Toán lớp 7 - Học toán với OnlineMath

Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?