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

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

Tương tự, ta có \(2b+ca=\left(b+c\right)\left(b+a\right)\)và \(2c+ab=\left(c+a\right)\left(c+b\right)\)

Vậy \(\left(2a+bc\right)\left(2b+ca\right)\left(2c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)là số chính phương.

Cách lầy nèk

\(Q=\frac{a}{1+2a}+\frac{b}{1+2b}\le\frac{a}{2\sqrt{2a}}+\frac{b}{2\sqrt{2b}}=\frac{\sqrt{a}}{2\sqrt{2}}+\frac{\sqrt{b}}{2\sqrt{2}}\)

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{2}}=\frac{\sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}}{2\sqrt{2}.\frac{1}{\sqrt{2}}}\le\frac{\frac{a+\frac{1}{2}}{2}+\frac{b+\frac{1}{2}}{2}}{2}=\frac{a+b+1}{4}=\frac{2}{4}=\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

Có a+b =1 

Áp dụng bất đẳng thức cô-si 

=> ab= \(\frac{\left(a+b\right)}{2}\) 

<=> ab= \(\frac{1}{2}\)

P=\(\frac{\left(ab+a+ab+b\right)}{\left(ab+a+b+1\right)}\)

= \(\frac{\left(2ab+1\right)}{\left(ab+2\right)}\)

=\(\frac{\left[\left(2ab+4\right)-3\right]}{\left(ab+2\right)}\)

=\(2+\left[\frac{-3}{\left(ab+2\right)}\right]\)
Có ab = \(\frac{1}{2}\)

\(ab+2\Leftarrow\frac{5}{2}\)

\(\frac{1}{\left(ab+2\right)}\ge\frac{2}{5}\)

\(\frac{-1}{\left(ab+2\right)}\Leftarrow\frac{-2}{5}\)

\(\frac{-3}{ \left(ab+2\right)}\Leftarrow\frac{-6}{5}\)

=> GTLN = \(\frac{-6}{5}+2=\frac{4}{5}\) tại \(a=b=\frac{1}{2}\)

Ta có : \(p=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(a+c\right)}+\frac{ab}{c^2\left(a+b\right)}\)

Áp dụng bất đẳng thức AM - GM ta có :

\(\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}.\frac{b+c}{4ab}}=\frac{1}{a}\)

\(\frac{ac}{b^2\left(a+c\right)}+\frac{a+c}{4ac}\ge4\sqrt{\frac{ac}{b^2\left(a+c\right)}.\frac{a+c}{4ac}}=\frac{1}{b}\)

\(\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}.\frac{a+b}{4ab}}=\frac{1}{c}\)

Cộng vế với vế ta được \(p+\frac{1}{4c}+\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4a}+\frac{1}{4c}+\frac{1}{4b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(\Leftrightarrow p+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(\Rightarrow p\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge3\sqrt[3]{\frac{1}{2a.2b.2c}}=\frac{3}{\sqrt[3]{8abc}}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Xét: \(\frac{bc}{a^2b+ca^2}=\frac{bc}{a\cdot abc\cdot\frac{1}{c}+a\cdot abc\cdot\frac{1}{b}}=\frac{b^2c^2}{ab+ca}\)(*)

Tương tự với (*) ta có: \(\hept{\begin{cases}\frac{ca}{b^2c+ab^2}=\frac{c^2a^2}{ab+bc}\\\frac{ab}{c^2a+bc^2}=\frac{a^2b^2}{ca+bc}\end{cases}}\)

\(\Rightarrow\Sigma_{cyc}\frac{bc}{a^2b+ca^2}=\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\)

Ta thấy\(\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\) có dạng: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\left(a+b+c\right)\)

Bước cuối Cô-si ba số và kết hợp điều kiện abc=1 là xong

a)

$3^x-y^3=1$

$\Leftrightarrow 3^x=y^3+1$

$\Leftrightarrow 3^x=(y+1)(y^2-y+1)$

$\gcd(y+1,y^2-y+1)=\gcd(y+1,3)$

Vì $3^x$ chỉ có ước nguyên tố là $3$ nên

$y+1=3^m,\quad y^2-y+1=3^n\qquad (m,n\in\mathbb N,\ m+n=x)$

Ta có $y^2-y+1-(y+1)(y-2)=3$ nên $\gcd(y+1,y^2-y+1)\mid 3$

Suy ra $\gcd(y+1,y^2-y+1)=1$ hoặc $3$.

Nếu $\gcd=1$ thì $y+1=1$

$\Rightarrow y=0$

$\Rightarrow 3^x=1$

$\Rightarrow x=0$.

Nếu $\gcd=3$ thì $3\mid y+1$

$\Rightarrow y\equiv2\pmod3$

$\Rightarrow y^2-y+1\equiv4-2+1\equiv3\equiv0\pmod3$

Lại có $y^2-y+1=(y+1)^2-3y$ nên $9\nmid (y^2-y+1)$

Suy ra $y^2-y+1=3$

$\Rightarrow y^2-y-2=0$

$\Rightarrow y=2$.

Khi đó $3^x=2^3+1=9$ $\Rightarrow x=2$.

Vậy $\boxed{(x,y)=(0,0)\ \text{hoặc}\ (2,2).}$

b/

$a+b+c=0$

$\Rightarrow c=-(a+b)$

$ab+2c^2=ab+2(a+b)^2$

$=2a^2+5ab+2b^2$

$=(2a+b)(a+2b)$

Tương tự $bc+2a^2=(2a+b)(a-b)$

$ca+2b^2=(a+2b)(b-a)$

Suy ra $(ab+2c^2)(bc+2a^2)(ca+2b^2)$

$=(2a+b)^2(a+2b)^2(a-b)(b-a)$

$=-(2a+b)^2(a+2b)^2(a-b)^2$ $\le 0$

Do đó $N=1-(ab+2c^2)(bc+2a^2)(ca+2b^2)$$=1+(2a+b)^2(a+2b)^2(a-b)^2$

$\ge 1$$>0$

=> N là số dương.