\(VT=\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ca}{c+a}+\dfrac{c\left(a+b+c\right)+ab}{a+b}\)

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

Ta có:

\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{c+a}\ge2\left(a+b\right)\)

Tương tự: \(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+c\right)\)

\(\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(b+c\right)\)

Cộng vế với vế:

\(\Rightarrow VT\ge2\left(a+b+c\right)=2\)

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

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}\\ \Leftrightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}b+c=2a\left(1\right)\\c+a=2b\left(2\right)\\a+b=2c\left(3\right)\end{matrix}\right.\\ \Leftrightarrow\left(1\right)-\left(2\right)=b-a=2a-2b\Leftrightarrow a-b=0\Leftrightarrow a=b\\ \left(2\right)-\left(3\right)=c-b=2b-2c\Leftrightarrow b-c=0\Leftrightarrow b=c\\ \left(3\right)-\left(1\right)=a-c=2c-2a\Leftrightarrow a-c=0\Leftrightarrow a=c\)

Vậy \(a=b=c\)

Giải:

Vì $a\in Z^{+}$

$\Rightarrow 5^b=a^3+3a^2+5>a+3=5^c$

$\Rightarrow 5^b>5^c\Rightarrow b>c$

$\Rightarrow 5^b⋮5^c$

$\Rightarrow a^3+3a^2+5⋮a+3$

$\Rightarrow a^2\left(a+3\right)+5⋮a+3$

Mà $a^2\left(a+3\right)⋮a+3$

$\Rightarrow 5⋮a+3$

$\Rightarrow a+3\in Ư\left(5\right)$

$\Rightarrow a+3\in \left\{\pm 1;\pm 5\right\}\left(1\right)$

Do $a\in Z^{+}\Rightarrow a+3\ge 4\left(2\right)$

Từ $\left(1\right)$ và $\left(2\right)$

$\Rightarrow a+3=5$

$\Rightarrow a=5-3$

$\Rightarrow a=2$$(\ast )$

Thay $(\ast )$ vào biểu thức ta có:

$2^3+{3.2}^2+5=5^b\iff b=2$

$2+3=5^c\iff c=1$

Vậy: $\{\begin{array}{c}a=2\\ b=2\\ c=1\end{array}$

Bạn xem lại xem viết đề đã đúng chưa vậy?

Để tìm Max M thì ta cần c/m \(a^2+b^2\le ab+1\)

Giả sử điều cần c/m là đúng , khi đó , ta có : 

\(a^2+b^2\le ab+1\)

\(\Leftrightarrow a^2-ab+b^2\le1\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\le a+b\)

\(\Leftrightarrow a^3+b^3\le a+b\)

\(\Leftrightarrow\left(a^3+b^3\right)^2\le\left(a+b\right)\left(a^5+b^5\right)\) ( do \(a^3+b^3=a^5+b^5\))

\(\Leftrightarrow a^6+2a^3b^3+b^6\le a^6+a^5b+b^5a+b^6\)

\(\Leftrightarrow2a^3b^3\le a^5b+b^5a\)

\(\Leftrightarrow a^5b+b^5a-2a^3b^3\ge0\)

\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)

\(\Leftrightarrow ab\left(a^2-b^2\right)\ge0\) ( điều này luôn đúng với a ; b dương ) 

=> Điều giả sử là đúng 

\(\Rightarrow a^2+b^2\le ab+1\)

\(\Rightarrow M=a^2+b^2-ab\le1\)

Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}ab=0\\a^2-b^2=0\end{cases}}\)  

\(\Leftrightarrow a=0\) hoặc \(b=0\)hoặc \(a^2=b^2\)

\(\Leftrightarrow a^2=b^2\)( a ,  b dương ) 

\(\Leftrightarrow a=b\)

Thế a = b vào b/t \(a^3+b^3=a^5+b^5\), ta có : 

\(2a^3=2a^5\)

\(\Leftrightarrow a^3=a^5\)\(\Leftrightarrow\frac{a^3}{a^5}=1\Leftrightarrow\frac{1}{a^2}=1\Leftrightarrow a=1\left(a>0\right)\)

\(\Leftrightarrow b=1\)

Vậy ...

Nguyen quang trung dung , trẻ trâu như mày quê tao đầy