no no sai r 

nghe da thay ngua r

Hệ PT trên \(< =>\hept{\begin{cases}2x^2y-2y^3=3x\\2x.\left(2x^2+2y^2\right)=20y\end{cases}}\)

\(< =>\hept{\begin{cases}2x^2y-2y^3=3x\\4xy^2+4x^3=20y\end{cases}}\)

\(< =>\hept{\begin{cases}2xy-2y^3=3\\4xy+4x^3=20\end{cases}}\)

\(< =>2xy+4x^3+2y^3=17\)

\(< =>2y\left(x+y^2\right)+4x^3=17\)

\(< =>2\left(yx+y^3+2x^3\right)=17\)

\(< =>y\left(x+y^2\right)+2x^3=\frac{17}{2}\)

\(< =>...\)

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

<=>  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab+bc+ca\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

<=>  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab+bc+ca\right)\ge9\)

Ap dung BDT AM-GM ta co:

\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab+bc+ca\right)\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+2\left(ab+bc+ca\right)\)

\(=\frac{3}{abc}+\left(ab+bc+ca\right)+\left(ab+bc+ca\right)\)

\(\ge3\sqrt[3]{\frac{3}{abc}\left(ab+bc+ca\right)\left(ab+bc+ca\right)}\)

\(\ge3\sqrt[3]{\frac{3}{abc}.3abc\left(a+b+c\right)}=9\)

=>  dpcm

Nếu để ý,bài này Cô si "ngược" là ra =))

Ta có: \(\sqrt{y-1}=\sqrt{1\left(y-1\right)}\le\frac{1+y-1}{2}=\frac{y}{2}\)

Tương tự: \(\sqrt{x-1}\le\frac{x}{2}\)

Do đó: \(x\sqrt{y-1}+y\sqrt{x-1}\le x.\frac{y}{2}+y.\frac{x}{2}=\frac{xy}{2}+\frac{xy}{2}=\frac{2xy}{2}=xy^{\left(đpcm\right)}\)

"=" xảy ra <=> y-1=1 và x-1=1 <=> x=y=2 (thỏa mãn)