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Đặt \(a=x,b=\frac{1}{x}\) thì ta có ab = 1
\(a-b=x-\frac{1}{x}=\frac{x^2-1}{x}=\frac{\left(x-1\right)\left(x+1\right)}{x}\). Vì \(x>1\) nên ta có \(a-b>0\)
\(3\left(a^2-b^2\right)< 2\left(a^3-b^3\right)\)
\(\Leftrightarrow3\left(a-b\right)\left(a+b\right)< 2\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow\left(a^2+ab+b^2\right)>\frac{3}{2}\left(a+b\right)\) (chia cả hai vế cho \(a-b>0\))
\(\Leftrightarrow\left(a^2-\frac{3}{2}a+\frac{9}{16}\right)+\left(b^2-\frac{3}{2}b+\frac{9}{16}\right)+\frac{7}{8}>0\)(vì ab = 1)
\(\Leftrightarrow\left(a-\frac{3}{4}\right)^2+\left(b-\frac{3}{4}\right)^2+\frac{7}{8}>0\) (luôn đúng)
Vậy có đpcm.
Đặt \(x=\left[x\right]+\left\{x\right\}\)
\(\Rightarrow\left[3x\right]=\left[3\left[x\right]+3\left\{x\right\}\right]=3\left[x\right]+\left[3\left\{x\right\}\right]\)
\(\left[x+\frac{2}{3}\right]=\left[\left[x\right]+\left\{x\right\}+\frac{2}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]\)
\(\left[x+\frac{1}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
\(\Rightarrow\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]=3\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
Ta cần chứng minh \(\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
- Nếu \(\frac{2}{3}\le\left\{x\right\}< 1\Rightarrow\left\{{}\begin{matrix}2\le\left[3\left\{x\right\}\right]< 3\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\1\le\left[\left\{x\right\}+\frac{1}{3}\right]< 2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=2\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=1\end{matrix}\right.\)
\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
- Nếu \(\frac{1}{3}\le\left\{x\right\}< \frac{2}{3}\Rightarrow\left\{{}\begin{matrix}1\le\left[3\left\{x\right\}\right]< 2\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\0\le\left[\left\{x\right\}+\frac{1}{3}\right]< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=1\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\)
\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
- Nếu \(0< \left\{x\right\}< \frac{1}{3}\) tương tự trên ta có:
\(\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=0\\\left[\left\{x\right\}+\frac{2}{3}\right]=0\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\) \(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)
\(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(P=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\left(\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(P=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-5x\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}\left(2+5x\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(P=\frac{-4x}{3-\sqrt{x}}\)
\(P=\frac{4x}{\sqrt{x}-3}\)
Có:
\(m\left(\sqrt{x}-3\right)P>x+1\)
\(\Leftrightarrow m\left(\sqrt{x}-3\right).\frac{4x}{\sqrt{x}-3}>x+1\)
\(\Leftrightarrow4mx>x+1\)
\(\Leftrightarrow4mx-x>1\)
\(\Leftrightarrow\left(4m-1\right)x>1\)
\(\Leftrightarrow x>\frac{1}{4m-1}\)
Lại có:
\(x>9\)
\(\Rightarrow\frac{1}{4m-1}< 9\)
\(\Leftrightarrow1< 9\left(4m-1\right)\)
\(\Leftrightarrow1< 36m-1\)
\(\Leftrightarrow10< 36m\)
\(\Leftrightarrow m< \frac{5}{18}\)
bài nào cx hỏi z :))
BĐT cần chứng minh tương đương với :
\(x+\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}+1>4\)
Áp dụng BĐT Cô-si,
Ta có : \(x+\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}+1=\frac{x^2-1}{x}+\frac{x+1}{2x}+\frac{x+1}{2x}+\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}\)
\(\ge4\)
Dấu "=" xảy ra khi \(\frac{x^2-1}{x}=\frac{x+1}{2x}=\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}\)
giải đc cái trên là vô nghiệm nên dấu "=" không xảy ra
bạn thanh tùng làm giống mình đó