Với \(n=4\) bđt \(\Leftrightarrow\)\(\frac{x_1}{x_4+x_2}+\frac{x_2}{x_1+x_3}+\frac{x_3}{x_2+x_4}+\frac{x_4}{x_3+x_1}\ge2\)

\(\Leftrightarrow\)\(\frac{x_1^2}{x_4x_1+x_1x_2}+\frac{x_2^2}{x_1x_2+x_2x_3}+\frac{x_3^2}{x_2x_3+x_3x_4}+\frac{x_4^2}{x_3x_4+x_4x_1}\ge2\) (1) 

\(VT_{\left(1\right)}\ge\frac{\left(x_1+x_2+x_3+x_4\right)^2}{2\left(x_1x_2+x_2x_3+x_3x_4+x_4x_1\right)}\ge\frac{\left(x_1+x_2+x_3+x_4\right)^2}{2.\frac{\left(x_1+x_2+x_3+x_4\right)^2}{4}}=2\)

Giả sử bđt đúng đến n=k hay \(\frac{x_1}{x_k+x_2}+\frac{x_2}{x_1+x_3}+...+\frac{x_{k-1}}{x_{k-2}+x_k}+\frac{x_k}{x_{k-1}+x_1}\ge2\)

\(\Leftrightarrow\)\(\frac{x_2}{x_1+x_3}+...+\frac{x_{k-1}}{x_{k-2}+x_k}\ge2-\frac{x_1}{x_k+x_2}-\frac{x_k}{x_{k-1}+x_1}\)

Với n=k+1, cần cm \(\frac{x_1}{x_{k+1}+x_2}+\frac{x_2}{x_1+x_3}+...+\frac{x_{k-1}}{x_{k-2}+x_k}+\frac{x_k}{x_{k-1}+x_{k+1}}+\frac{x_{k+1}}{x_k+x_1}\ge2\)

hay \(\frac{x_1}{x_{k+1}+x_2}-\frac{x_1}{x_k+x_2}+\frac{x_k}{x_{k-1}+x_{k+1}}-\frac{x_k}{x_{k-1}+x_1}+\frac{x_{k+1}}{x_k+x_1}\ge0\) (2) 

giả sử \(x_k=max\left\{a_1;a_2;...;a_{k+1}\right\}\)

\(VT_{\left(2\right)}=\frac{x_1\left(x_k-x_{k+1}\right)}{\left(x_k+x_2\right)\left(x_{k+1}+x_2\right)}+\frac{x_k\left(x_1-x_{k+1}\right)}{\left(x_{k-1}+x_1\right)\left(x_{k-1}+x_{k+1}\right)}+\frac{x_{k+1}}{x_k+x_1}>0\)

nhầm, chỗ giả sử là \(x_{k+1}=min\left\{x_1;x_2;...;x_{k+1}\right\}\)

*)Giả sử với \(n=2\) đặt \(\hept{\begin{cases}2x=b+c-a\\2y=a-b+c\\2z=a+b-c\end{cases}\left(x,y,z>0\right)}\)

\(\Rightarrow a=y+z;b=x+z;c=x+y\)

BĐT cần chứng minh là \(xy^3+yz^3+xz^3-xyz\left(x+y+z\right)\ge0\)

Tự C/M cái này bằng AM-GM nhé

*)Giả sử đúng với n (tức là dạng t/q). KO mất tính tổng quát giả sử \(a\ge b\ge c\)

Khi đó ta có: \(b^nc(b-c)\ge-a^nb(a-b)-c^na(c-a)\)

\(\Rightarrow b^{n+1}c(b-c)\ge-a^nb^2(a-b)-c^nab(c-a)\)

Nên \(a^{n+1}b(a-b)+b^{n+1}c(b-c)+c^{n+1}a(c-a)\)

\(\ge a^{n+1}b(a-b)-a^nb^2(a-b)-c^nab(c-a)+c^{n+1}a(c-a)\)

\(=a^nb(a-b)+b^nc(b-c)+c^na(c-a)\ge0\) 

Theo nguyên lí quy nạp thì có ĐPCM

\(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}6x-1=5\\6x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

Ai trên 10 thì mình nha mình lại 3 luôn

ta có\

\(\left(\sqrt{x}+2\sqrt{y}\right)^2\subseteq\left(1^2+2^2\right)\left(x+y\right)\)

\(< =>10^2\subseteq5\left(x+y\right)\)

\(< =>20\subseteq x+y\)

chết mik làm rồi ra v

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