1. Áp dụng Min - cốp - ski, ta được: \(\sqrt{\frac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\frac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\frac{9}{\left(c+a\right)^2}+b^2}\)\(\ge\sqrt{\left(\frac{3}{a+b}+\frac{3}{b+c}+\frac{3}{c+a}\right)^2+\left(a+b+c\right)^2}\)\(\ge\sqrt{\left(\frac{27}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)(Bunyakovsky dạng phân thức)

Đặt \(t=a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)thì ta cần chứng minh: \(\sqrt{\frac{729}{4t^2}+t^2}\ge\frac{3\sqrt{13}}{2}\Leftrightarrow\frac{729}{4t^2}+t^2\ge\frac{117}{4}\)\(\Leftrightarrow\frac{\left(t+3\right)\left(t-3\right)\left(2t+9\right)\left(2t-9\right)}{4t^2}\ge0\)*đúng bởi \(t-3\le0;t+3>0;2t+9>0;2t-9< 0;4t^2>0\)*

Đẳng thức xảy ra khi t = 3 hay a = b = c = 1

2. Ta có: \(\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}-3=\frac{\left(x^2-y^2\right)^2\left(x^4+y^4+x^2y^2\right)}{x^2y^2\left(x^2+y^2\right)^2}\ge0\)\(\Rightarrow\frac{4x^2y^2}{\left(x^2+y^2\right)^2}+\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge3\)

Đẳng thức xảy ra khi x = y

9999.55553=555474447

Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)

=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)

=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)

=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=> \(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=> \(\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}=\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\)

=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)

=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)

=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)

=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)

x + y =1 nên x + yt = 2

t + b= = dr wn e=gv

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Xét hệ \(\hept{\begin{cases}x^2+xy^2-xy-y^3=0\left(1\right)\\x^2-3y^2+2\sqrt{x}=0\left(2\right)\end{cases}}\):

\(\left(1\right)\Leftrightarrow\left(x^2-xy\right)+\left(xy^2-y^3\right)=0\Leftrightarrow x\left(x-y\right)+y^2\left(x-y\right)=0\)\(\Leftrightarrow\left(x+y^2\right)\left(x-y\right)=0\Leftrightarrow\orbr{\begin{cases}y^2=-x\\x=y\end{cases}}\)

Trường hợp 1: \(y^2=-x\)thay vào (2), ta được: \(x^2+3x+2\sqrt{x}=0\)(*)

Đặt \(\sqrt{x}=t\left(t>0\right)\)thì (*) trở thành: \(t^4+3t^2+2t=0\Leftrightarrow t\left(t^3+3t+2\right)=0\Leftrightarrow t=0\)(Lưu ý: Phương trình \(t^3+3t+2=0\)có một nghiệm âm và hai nghiệm phức đều không thỏa mãn)

\(\Rightarrow x=y=0\)

Trường hợp 2: \(x=y\)thay vào (2), ta được: \(x^2-3x^2+2\sqrt{x}=0\Leftrightarrow2\sqrt{x}-2x^2=0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x^3}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\Rightarrow y=0\\x=1\Rightarrow y=1\end{cases}}\)

Vậy hệ có 2 nghiệm (x,y) = {(1;1) ; (0;0)}