Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+a+b+c=2+2018\)

\(\Leftrightarrow\frac{a+ab+bc}{b+c}+\frac{b+bc+ab}{c+a}+\frac{c+ac+bc}{a+b}=2020\)

\(\Leftrightarrow a\left(\frac{1+b+c}{b+c}\right)+b\left(\frac{1+a+c}{a+c}\right)+c\left(\frac{1+a+b}{a+b}\right)=2020\left(1\right)\)

Vì \(a+b+c=2018\Rightarrow\hept{\begin{cases}a+b=2018-c\\b+c=2018-a\\c+a=2018-b\end{cases}\left(2\right)}\)

Thay (2) vào (1) ta được: 

\(a\left(\frac{2019-a}{b+c}\right)+b\left(\frac{2019-b}{a+c}\right)+c\left(\frac{2019-c}{a+b}\right)=2020\)

\(\Leftrightarrow\frac{2019a-a^2}{b+c}+\frac{2019b-b^2}{a+c}+\frac{2019c-c^2}{a+b}=2020\)

\(\Leftrightarrow\frac{2019a}{b+c}-\frac{a^2}{b+c}+\frac{2019b}{a+c}-\frac{b^2}{a+c}+\frac{2019c}{a+b}-\frac{c^2}{a+b}=2020\)

\(\Leftrightarrow2019\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)

\(\Leftrightarrow4038-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)( vì \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\))

\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=2018\)

\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+1=2019\)

neu de bai bai 1 la tinh x+y thi mik lam cho

đăng từng này thì ai làm cho 

We have \(P=\frac{x^4+2x^2+2}{x^2+1}\)

\(\Rightarrow P=\frac{x^4+2x^2+1+1}{x^2+1}\)

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

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

\(\ge2\sqrt{\frac{x^2+1}{x^2+1}}=2\)

(Dấu "="\(\Leftrightarrow x=0\))

Vậy \(P_{min}=2\Leftrightarrow x=0\)

\(K=\frac{a^2}{c\left(a^2+c^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\left(a,b,c>0\right)\).

Ta có:

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

Vì \(a,c>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(a^2+c^2\ge2ac\).

\(\Leftrightarrow c\left(a^2+c^2\right)\ge2ac^2\).

\(\Rightarrow\frac{1}{c\left(a^2+c^2\right)}\le\frac{1}{2ac^2}\)

\(\Leftrightarrow\frac{c^2}{c\left(a^2+c^2\right)}\le\frac{c^2}{2ac^2}=\frac{1}{2a}\).

\(\Leftrightarrow-\frac{c^2}{c\left(a^2+c^2\right)}\ge-\frac{1}{2a}\).

\(\Leftrightarrow\frac{1}{c}-\frac{c^2}{c\left(a^2+c^2\right)}\ge\frac{1}{c}-\frac{1}{2a}\)

\(\Leftrightarrow\frac{a^2}{c\left(a^2+c^2\right)}\ge\frac{1}{c}-\frac{1}{2a}\left(1\right)\)

Dấu bằng xảy ra \(\Leftrightarrow a=c>0\) .

Chứng minh tương tự, ta được:

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

Dấu bằng xảy ra \(\Leftrightarrow a=b>0\)

Chứng minh tương tự, ta dược:

\(\frac{c^2}{b\left(b^2+c^2\right)}\ge\frac{1}{b}-\frac{1}{2c}\left(b,c>0\right)\left(3\right)\).

Dấu bằng xảy ra \(\Leftrightarrow b=c>0\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(\frac{a^2}{c\left(a^2+c^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}\ge\)\(\frac{1}{c}-\frac{1}{2a}+\frac{1}{a}-\frac{1}{2b}+\frac{1}{b}-\frac{1}{2c}\).

\(\Leftrightarrow K\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).

\(\Leftrightarrow K\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\).

\(\Leftrightarrow K\ge\frac{1}{2}\left(\frac{ab+bc+ca}{abc}\right)\).

Mà \(ab+bc+ca=3abc\)(theo đề bài).

Do đó \(K\ge\frac{1}{2}.\frac{3abc}{abc}\).

\(\Leftrightarrow K\ge\frac{3abc}{2abc}\).

\(\Leftrightarrow K\ge\frac{3}{2}\).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=3abc\end{cases}}\Leftrightarrow a=b=c=1\).

Vậy \(minK=\frac{3}{2}\Leftrightarrow a=b=c=1\).

Đặt \(A=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)

Không mất tính tổng quát giả sử \(a\ge b\ge c\ge d\)=>\(a^2\ge b^2\ge c^2\ge d^2\)

=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\le\frac{1}{d^2}\)

=>\(A\le\frac{4}{d^2}\)=>\(d^2\le4\)=>\(d\in\text{ }\text{{}\pm1,\pm2\text{ }\)

Xét \(d=\pm1\)=> vô lí

Xét d=\(\pm\)2=> a=b=c=d=\(\pm\)2

=> M=ab+cd=4+4=8

co nhieu cau tuong tu tren mang ban tu tm hieu nhe