giải xy+z^2=yz+x^2=xz+y^2 cho ta x=y=z nên x=y=z

= cộng trừ căn 3

cộng trừ căn 2/3

Ta có:\(a+b+c=1\Leftrightarrow1=\left[\left(a+b\right)+c\right]^2\ge4\left(a+b\right)c\)

\(\Rightarrow\frac{a+b}{abc}\ge\frac{4\left(a+b\right)^2c}{abc}\ge\frac{4.4ab.c}{abc}=16\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}a=b\\a+b=c\\a+b+c=1\end{cases}\Leftrightarrow\hept{\begin{cases}a=b=\frac{1}{4}\\c=\frac{1}{2}\end{cases}}}\)

Vậy P đạt GTNN là 16 khi và chỉ khi a=b=1/4 và c=1/2

\(5x^2+4y^2-8xy+4x-8y+3=0\)

\(\Leftrightarrow4x^2+4y^2-2.2x.2y+2.2x.2-2.2y.2+4+x^2-4x+4=5\)

\(\Leftrightarrow\left(2x-2y+2\right)^2+\left(x-2\right)^2=5=\left(\pm1\right)^2+\left(\pm2\right)^2\)

Ta có các trường hợp sau: 

- \(\hept{\begin{cases}2x-2y+2=2\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)

- \(\hept{\begin{cases}2x-2y+2=2\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}\)

- \(\hept{\begin{cases}2x-2y+2=-2\\x-2=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)

- \(\hept{\begin{cases}2x-2y+2=-2\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=5\end{cases}}\)

\(Q=\frac{c+ab}{a+b}+...+\frac{b+ac}{a+c};\frac{c+ab}{a+b}=\frac{ca+cb+c^2+ab}{a+b}=\frac{\left(c+b\right)\left(c+a\right)}{a+b}\)

\(\text{tương tự ta có:}2Q=\frac{2\left(a+b\right)\left(b+c\right)}{a+c}+\frac{2\left(b+c\right)\left(a+c\right)}{a+b}+\frac{2\left(a+b\right)\left(a+c\right)}{b+c}\)

\(\ge2\left(\sqrt{\frac{\left(a+b\right)^2\left(b+c\right)\left(c+a\right)}{\left(b+c\right)\left(c+a\right)}}+\sqrt{\frac{\left(b+c\right)^2\left(a+c\right)\left(a+b\right)}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\frac{\left(c+a\right)^2\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)}}\right)\)

\(=2\left[2\left(a+b+c\right)\right]=4\Rightarrowđpcm\text{ dấu "=":}a=b=c=\frac{1}{3}\)