Ta có : Hệ \(\hept{\begin{cases}x^3+xy^2-10y=0\\x^2+6y^2=10\end{cases}}\Leftrightarrow\hept{\begin{cases}x\left(x^2+y^2\right)-10y=0\\x^2+6y^2=10\end{cases}}\) 

\(\Leftrightarrow\)\(\hept{\begin{cases}x\left(10-6y^2+y^2\right)-10y=0\\x^2=10-6y^2\end{cases}\Leftrightarrow\hept{\begin{cases}2x-xy^2-2y=0\\x^2=10-6y^2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{2y}{2-y^2}\\x^2=10-6y^2\end{cases}}\)(1) 

Với y = \(\pm\sqrt{2}\)=> \(∄\)x thỏa mãn hệ 

=> y \(\ne\pm\sqrt{2}\)

Khi đó hệ (1) <=> \(\hept{\begin{cases}\left(\frac{2y}{2-y^2}\right)^2=10-6y^2\\x^2=10-6y^2\end{cases}}\Leftrightarrow\hept{\begin{cases}6y^6-34y^4+68y^2-40=0\left(2\right)\\x^2=10-6y^2\left(^∗\right)\end{cases}}\)

Đặt t = y² \(\ge0\)

Khi đó (2) <=> 6t³ - 34t² + 68t - 40 = 0

<=> 3t³ - 17t² + 34t - 20 = 0

<=> (3t³ - 3) - 17(t² - 2t + 1) = 0

<=> 3(t - 1)(t² + t + 1) - 17(t - 1)² = 0

<=> (t - 1)(3t² - 14t + 20) = 0

<=> t - 1 = 0 (Vì 3t² - 14t + 20 > 0 \(\forall t\)) 

<=> t = 1

Khi đó y² = 1 <=> y = \(\pm1\)

Thay y = \(\pm1\)vào (*) 

=> x² = 10 - 6y² = 10 - 6 = 4 <=> x = \(\pm2\)
Vậy hệ có 4 nghiệm (2 ; 1) ; (2 ; - 1) ; (-2 ; - 1) ; (-2 ; 1) 

\(\Rightarrow x^3+xy^2-\left(x^2+6y^2\right)y=0\)

\(\Leftrightarrow x^3-x^2y+xy^2-6y^3=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x^2+xy+3y^2\right)=0\)

\(\Rightarrow x=2y\)

Thế vào \(x^2+6y^2=10\)

\(\Rightarrow10y^2=10\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-1\Rightarrow x=-2\end{matrix}\right.\)

ĐKXĐ: \(x\ne-1;y\ne2\)

\(\left\{{}\begin{matrix}\dfrac{3}{x+1}-\dfrac{4}{y-2}+1=0\\x+1+y-2=2\left(x+1\right)\left(y-2\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{x+1}-\dfrac{4}{y-2}=-1\\\dfrac{1}{x+1}+\dfrac{1}{y-2}=2\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+1}=u\\\dfrac{1}{y-2}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3u-4v=-1\\u+v=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=1\\\dfrac{1}{y-2}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+1=1\\y-2=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=3\end{matrix}\right.\)

\(\left(a^2+\dfrac{1}{b^2}\right)\left(\dfrac{1}{4}+4\right)\ge\left(\dfrac{a}{2}+\dfrac{2}{b}\right)^2\)

\(\Rightarrow\sqrt{a^2+\dfrac{1}{b^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{a}{2}+\dfrac{2}{b}\right)\)

Tương tự: \(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{b}{2}+\dfrac{2}{c}\right)\) ; \(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{c}{2}+\dfrac{2}{a}\right)\)

Cộng vế:

\(S\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{a+b+c}{2}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{a+b+c}{2}+\dfrac{18}{a+b+c}\right)\)

\(S\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{a+b+c}{2}+\dfrac{9}{8\left(a+b+c\right)}+\dfrac{135}{8\left(a+b+c\right)}\right)\)

\(S\ge\dfrac{2}{\sqrt{17}}\left(2\sqrt{\dfrac{9\left(a+b+c\right)}{16\left(a+b+c\right)}}+\dfrac{135}{8.\dfrac{3}{2}}\right)=\dfrac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)

Chỗ này mình còn chưa hiểu bạn giải thích giúp mình với  \(S\ge\frac{2}{17}\left(\frac{a+b+c}{2}+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\right)\ge\frac{2}{17}\left(\frac{a+b+c}{2}+\frac{18}{a+b+c}\right)\)

Cách 1:

Do vai trò của a;b;c là như nhau, không mất tính tổng quát, giả sử \(a\ge b\ge c\)

\(\Rightarrow3=ab+bc+ca\le3ab\Rightarrow ab\ge1\)

Ta có:

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}=\dfrac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}=1-\dfrac{a^2b^2-1}{a^2b^2+a^2+b^2+1}\)

\(\ge1-\dfrac{a^2b^2-1}{a^2b^2+2ab+1}=1-\dfrac{ab-1}{ab+1}=\dfrac{2}{1+ab}\)

\(\Rightarrow VT\ge\dfrac{2}{1+ab}+\dfrac{1}{1+c^2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{2}{1+ab}+\dfrac{1}{1+c^2}\ge\dfrac{3}{2}\Leftrightarrow c^2+3-ab\ge3abc^2\)

\(\Leftrightarrow c^2+ac+bc\ge3abc^2\Leftrightarrow a+b+c\ge3abc\)

\(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge3\)

Đúng do \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{9}{ab+bc+ca}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Cách 2:

\(\Leftrightarrow1-\dfrac{a^2}{a^2+1}+1-\dfrac{b^2}{b^2+1}+1-\dfrac{c^2}{c^2+1}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{3a^2}{3a^2+3}+\dfrac{3b^2}{3b^2+3}+\dfrac{3c^2}{3c^2+3}\le\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{3a^2}{2a^2+a^2+ab+bc+ca}+\dfrac{3b^2}{2b^2+b^2+ab+bc+ca}+\dfrac{3c^2}{2c^2+c^2+ab+bc+ca}\le\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{a^2}{a\left(a+b+c\right)+2a^2+bc}+\dfrac{b^2}{b\left(a+b+c\right)+2b^2+ac}+\dfrac{c^2}{c\left(a+b+c\right)+2c^2+ab}\le\dfrac{1}{2}\)

Ta có:

\(\dfrac{a^2}{a\left(a+b+c\right)+2a^2+bc}\le\dfrac{1}{4}\left(\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{2a^2+bc}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{4}\left(1+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)

Nên ta chỉ cần chứng minh:

\(\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\le1\)

\(\Leftrightarrow\dfrac{bc}{2a^2+bc}+\dfrac{ac}{2b^2+ac}+\dfrac{ab}{2c^2+ab}\ge1\)

\(\Leftrightarrow\dfrac{\left(bc\right)^2}{2a^2bc+\left(bc\right)^2}+\dfrac{\left(ca\right)^2}{2ab^2c+\left(ac\right)^2}+\dfrac{\left(ab\right)^2}{2abc^2+\left(ab\right)^2}\ge1\)

Đúng do:

\(VT\ge\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\)

Gọi I là tâm đường tròn nội tiếp EOF, C và D lần lượt là tiếp điểm của (I) với OE và OF

Tứ giác ICOD là hình chữ nhật (có 3 góc vuông)

Mà \(IC=ID=r\Rightarrow ICOD\) là hình vuông

\(S_{IEF}+S_{IEO}+S_{IFO}=\dfrac{1}{2}\left(IG.EF+IC.EO+ID.FO\right)\)

\(=\dfrac{1}{2}r\left(EF+EO+FO\right)\) (do \(IG=IC=ID=r\))

\(=S_{OEF}=\dfrac{1}{2}OM.EF=\dfrac{1}{2}R.EF\)

\(\Rightarrow\dfrac{r}{R}=\dfrac{EF}{EF+OE+OF}>\dfrac{EF}{EF+EF+EF}=\dfrac{1}{3}\)

(do tam giác OEF vuông nên \(OE< EF;OF< EF\))