Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a+b\ge2\sqrt{ab}\Leftrightarrow2\sqrt{ab}\le4\Leftrightarrow ab\le4\)
\(P=\left(\dfrac{2}{a^2+b^2}+\dfrac{1}{ab}\right)+\dfrac{2}{ab}+2ab+\dfrac{32}{ab}\\ \Leftrightarrow P=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{2}{ab}+2ab+\dfrac{32}{ab}\\ \Leftrightarrow P\ge2\cdot\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{32}{ab}\cdot2ab}+\dfrac{2}{4}\\ \Leftrightarrow P\ge\dfrac{8}{\left(a+b\right)^2}+2\sqrt{64}+\dfrac{1}{2}\\ \Leftrightarrow P\ge\dfrac{8}{16}+16+\dfrac{1}{2}=17\)
Dấu \("="\Leftrightarrow a=b=2\)
\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)
\(=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{34}{ab}+\dfrac{17}{8}ab-\dfrac{1}{8}ab\)
\(\ge2.\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{34}{ab}.\dfrac{17}{8}ab}-\dfrac{1}{8}.\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow A\ge2.\dfrac{4}{\left(a+b\right)^2}+2.\dfrac{17}{2}-\dfrac{1}{8}.\dfrac{4^2}{4}\ge2.\dfrac{4}{4^2}+17-\dfrac{1}{2}\)
\(\Leftrightarrow A\ge\dfrac{1}{2}+17-\dfrac{1}{2}=17\)
Dấu "=" <=> a = b = 2
\(=\left(1^2+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)\ge\left(1a+4.\dfrac{1}{b}\right)^2\\ \Rightarrow\sqrt{a^2+\dfrac{1}{vb^2}}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\)
Tương tự
\(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\\ \sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\\ Do.đó:\\ Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\ge\dfrac{1}{\sqrt{17}}\\ \left(a+b+c+\dfrac{36}{a+b+c}\right)\)
\(=\dfrac{1}{\sqrt{17}}\\ \left[a+b+c+\dfrac{9}{4\left(a+b+c\right)}+\dfrac{135}{4\left(a+b+c\right)}\right]\\ \ge\dfrac{3\sqrt{17}}{2}\)
Cái thứ nhất là tại sao có cái đầu tiên =)) cái thứ 2 dấu bằng xảy ra khi nào :V
\(Q=\dfrac{2-\dfrac{c}{a}-\dfrac{2b}{a}+\left(\dfrac{b}{a}\right)\left(\dfrac{c}{a}\right)}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2-mn+2\left(m+n\right)-mn\left(m+n\right)}{1+m+n+mn}\)
\(Q=\dfrac{\left(2-mn\right)\left(m+n+1\right)}{\left(m+1\right)\left(n+1\right)}\ge\dfrac{\left[8-\left(m+n\right)^2\right]\left(m+n+1\right)}{\left(m+n+2\right)^2}\)
Đặt \(m+n=t\Rightarrow0\le t\le2\)
\(Q\ge\dfrac{\left(8-t^2\right)\left(t+1\right)}{\left(t+2\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{\left(2-t\right)\left(4t^2+15t+10\right)}{4\left(t+2\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(t=2\) hay \(m=n=1\)
Thầy ơi sao bên này là (2-mn) qua bên kia lại là \(\left[8-\left(m+n\right)^2\right]\) , dưới mẫu là (m+1)(n+1) qua bên này là \(\text{(m+n+2)}^2\)
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
Áp dụng BĐT Svac
⇒\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\text{≥}\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)
Vì a+b+c=6
⇒\(\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{6^2}{12}=\dfrac{36}{12}=3\)
Còn lại thì bạn tự làm tiếp nha
\(không\) \(dùng\) \(bđt\) \(làm\) \(sao\) \(ra\) \(được\) ??
\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}.\sqrt{\left(1+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(bunhiacopki\right)\)
\(tương-tự:\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\)
\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\)
\(\Rightarrow Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)
\(bđt:cosi\Rightarrow16a+\dfrac{4}{a}\ge2\sqrt{16a.\dfrac{4}{a}}=2\sqrt{16.4}=16\)
\(tương-tự\Rightarrow16b+\dfrac{4}{b}\ge16;16c+\dfrac{4}{c}\ge16\)
\(có:a+b+c\le\dfrac{3}{2}\Rightarrow15\left(a+b+c\right)\le\dfrac{45}{2}\)
\(\Rightarrow-15\left(a+b+c\right)\ge-\dfrac{45}{2}\)
\(\Rightarrow Q\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)
\(dấu"="xayra\Leftrightarrow a=b=c=\dfrac{1}{2}\)
các bước ban đầu dùng bunhia chọn được 1+4^2 là do dự đoán được trước điểm rơi tại a=b=c=1/2 thôi bạn,cả bước tách dùng cosi cũng dự đoán dc điểm rơi =1/2 nên tách đc thôi
Tại sao lại k được dùng nhỉ? Trông khi dùng thì bài toán sẽ dễ giải quyết hơn
Áp dụng Bunhiacopxki:
\(\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(\dfrac{1}{4}+4\right)}\ge\dfrac{a}{2}+\dfrac{2}{b}\)
\(\Rightarrow\sqrt{a^2+\dfrac{1}{b^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{a}{2}+\dfrac{2}{b}\right)\)
Do đó:
\(Q\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{a+b+c}{2}+2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)
\(\Rightarrow Q\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{a+b+c}{2}+\dfrac{18}{a+b+c}\right]\)
Áp dụng Cô-si:
\(\dfrac{a+b+c}{2}+\dfrac{9}{8\left(a+b+c\right)}\ge\dfrac{3}{2}\)
Do đó:
\(Q\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{3}{2}+\dfrac{135}{8\left(a+b+c\right)}\right]\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{3}{2}+\dfrac{135}{8.\dfrac{3}{2}}\right]=\dfrac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
\(A=a^2+\dfrac{1}{16a^2}+b^2+\dfrac{1}{16b^2}+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(A\ge2\sqrt{\dfrac{a^2}{16a^2}}+2\sqrt{\dfrac{b^2}{16b^2}}+\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)
\(A\ge1+\dfrac{15}{32}\left(\dfrac{4}{a+b}\right)^2\ge1+\dfrac{15}{32}.4\)
Bài 1:
dự đoán dấu = sẽ là \(a^2=b^2=c^2=\dfrac{1}{2}\) nên cứ thế mà chém thôi .
Ta có: \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\)
Bunyakovsky:\(\left(a^2+\dfrac{1}{2}\right)\left(\dfrac{1}{2}+b^2\right)+\dfrac{1}{2}\left(a^2+b^2\right)+\dfrac{3}{4}\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\)
\(VT=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{3}{4}\left[\left(a+b\right)^2+1\right]\left(1+c^2\right)\ge\dfrac{3}{4}\left(a+b+c\right)^2\)(đpcm)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{2}}\)
P/s: còn 1 cách khác nữa đó là khai triển sau đó xài schur . Chi tiết trong tệp BĐT schur .pdf
\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\\ =\dfrac{2}{a^2+b^2}+\dfrac{2}{2ab}+\dfrac{34}{ab}+\dfrac{17ab}{8}-\dfrac{ab}{8}\\ =2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+17\left(\dfrac{2}{ab}+\dfrac{ab}{8}\right)-\dfrac{ab}{8}\\ \overset{AM-GM}{\ge}2\cdot\dfrac{1}{a^2+b^2+2ab}+17\sqrt{\dfrac{2}{ab}\cdot\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{2}{\left(a+b\right)^2}+\dfrac{17}{2}-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{2}{4^2}+\dfrac{17}{2}-\dfrac{4^2}{32}=\dfrac{65}{8}\)
Dấu "=" xảy ra khi : \(\left\{{}\begin{matrix}\dfrac{2}{ab}=\dfrac{ab}{8}\\a^2+b^2=2ab\\a=b\\a+b=4\end{matrix}\right.\Leftrightarrow a=b=2\)
Vậy \(A_{Min}=\dfrac{65}{8}\) khi \(a=b=2\)