Cho: \(\left\{{}\begin{matrix}a,b,c>0\\a^2+b^2+c^2=abc\end{matrix}\right.\)
Tìm Max:
\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ac}+\dfrac{c}{c^2+ab}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2022
\(P=\dfrac{2-\left(1+a^2\right)}{1+a^2}+\dfrac{2-\left(1+b^2\right)}{1+b^2}+\dfrac{2}{\sqrt{1+c^2}}\)
\(P=2\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}\right)-2\)
Từ điều kiện \(ab+bc+ca=1\), đặt \(\left\{{}\begin{matrix}a=tanx\\b=tany\\c=tanz\end{matrix}\right.\) với \(x+y+z=\dfrac{\pi}{2}\)
Xét \(Q=\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}=\dfrac{1}{1+tan^2x}+\dfrac{1}{1+tan^2y}+\dfrac{1}{\sqrt{1+tan^2z}}\)
\(Q=cos^2x+cos^2y+cosz=1+\dfrac{1}{2}\left(cos2x+cos2y\right)+cosz\)
\(=1+cos\left(x+y\right)cos\left(x-y\right)+cosz\le1+cos\left(x+y\right)+cosz\)
\(=1+cos\left(\dfrac{\pi}{2}-z\right)+cosz=1+sinz+cosz=1+\sqrt{2}sin\left(z+\dfrac{\pi}{4}\right)\le1+\sqrt{2}\)
\(\Rightarrow P\le2\left(1+\sqrt{2}\right)-2=2\sqrt{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=y=\dfrac{\pi}{8}\\z=\dfrac{\pi}{4}\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(\sqrt{2}-1;\sqrt{2}-1;1\right)\)
14 tháng 2 2021
Lâu rồi không lên Hoc24
Áp dụng bất đẳng thức Minkowski, Schwarz và AM - GM ta có:
\(S\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}=\sqrt{\left[\left(a+b+c\right)^2+\dfrac{81}{16\left(a+b+c\right)^2}\right]+\dfrac{81.15}{16\left(a+b+c\right)^2}}\ge\sqrt{\dfrac{9}{2}+\dfrac{135}{4}}=\sqrt{\dfrac{153}{4}}=\dfrac{3\sqrt{17}}{2}\).
14 tháng 2 2021
Sau khi chọn đc hệ số điểm rơi là 16 thì cơ sở nào tách tiếp ra 16 số rồi áp dụng cosi nữa vậy ạ??
15 tháng 4 2023
+) Bài bất đẳng thức:
\(\dfrac{2017a-a^2}{bc}=\dfrac{\left(a+b+c\right)a-a^2}{bc}=\dfrac{ab+ca}{bc}=\dfrac{a}{c}+\dfrac{a}{b}\left(1\right)\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{2017b-b^2}{ca}=\dfrac{b}{a}+\dfrac{b}{c}\left(2\right)\\\dfrac{2017c-c^2}{ab}=\dfrac{c}{a}+\dfrac{c}{b}\left(3\right)\end{matrix}\right.\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow\dfrac{2017a-a^2}{bc}+\dfrac{2017b-b^2}{bc}+\dfrac{2017c-c^2}{ab}=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\)
\(\sqrt{2}\left(\sum\sqrt{\dfrac{2017-a}{a}}\right)=\sqrt{2}\left(\sum\sqrt{\dfrac{\left(a+b+c\right)-a}{a}}\right)=\sqrt{2}\left(\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}+\sqrt{\dfrac{a+b}{2}}\right)\)
Bất đẳng thức cần chứng minh tương đương với:
\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge\sqrt{2}\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)\)
*Có: \(\sqrt{2.\dfrac{a+b}{c}}+\sqrt{2.\dfrac{b+c}{a}}+\sqrt{2.\dfrac{c+a}{b}}\le\dfrac{2+\dfrac{a+b}{c}}{2}+\dfrac{2+\dfrac{b+c}{a}}{2}+\dfrac{2+\dfrac{c+a}{b}}{2}=3+\dfrac{\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}}{2}\)
Ta chỉ cần chứng minh:
\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge3+\dfrac{\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}}{2}\)
hay \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\) (cái này chị tự chứng minh nhé)
HN
12 tháng 5 2017
Đặt \(\dfrac{b}{c}=x\)
Ta có: \(\left\{{}\begin{matrix}ab+bc=2c^2\\2a\le c\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{c}.x+x=2\\\dfrac{a}{c}\le\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{2-x}{x}\\\dfrac{2-x}{x}\le\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{2-x}{x}\\x\ge\dfrac{4}{3}\end{matrix}\right.\)
Ta lại có:
\(\dfrac{a}{a-b}+\dfrac{b}{b-c}+\dfrac{c}{c-a}=\dfrac{\dfrac{a}{c}}{\dfrac{a}{c}-\dfrac{b}{c}}+\dfrac{\dfrac{b}{c}}{\dfrac{b}{c}-1}+\dfrac{1}{1-\dfrac{a}{c}}\)
\(=\dfrac{\dfrac{2-x}{x}}{\dfrac{2-x}{x}-x}+\dfrac{x}{x-1}+\dfrac{1}{1-\dfrac{2-x}{x}}\)
\(=\dfrac{3x^2+8x-4}{2x^2+2x-4}\)
\(=\dfrac{27}{5}+\dfrac{39x^2+14x-88}{2x^2+2x-4}=\dfrac{27}{5}+\dfrac{\left(3x-4\right)\left(13x+22\right)}{2\left(x-1\right)\left(x+2\right)}\ge\dfrac{27}{5}\)
Vậy GTNN là \(\dfrac{27}{5}\) dấu = xảy ra khi \(x=\dfrac{4}{3}\)
4 tháng 11 2017
Áp dụng BĐT B.C.S ta có
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}\)
mặt khác do \(a+b+c\le3\Rightarrow\dfrac{9}{\left(a+b+c\right)^2}\ge1\)
\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge1\)(*)
ta lại có \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}\le3\)
\(\Rightarrow\dfrac{2007}{ab+bc+ac}\ge\dfrac{2007}{3}=669\)(**)
lấy (*)+(**) vế theo vế ta được
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ac}\ge669+1=670\left(dpcm\right)\)
MD
22 tháng 4 2018
\(\Leftrightarrow y=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)
Ta có: \(\dfrac{\sqrt{c-2}}{c}\le\dfrac{1}{2\sqrt{2}}\Leftrightarrow\left(\sqrt{c-2}-\sqrt{2}\right)^2\ge0\) ( Luôn đúng)
Tương tự: \(\dfrac{\sqrt{a-3}}{a}\le\dfrac{1}{2\sqrt{3}};\dfrac{\sqrt{b-4}}{b}\le\dfrac{1}{4}\)
\(\Rightarrow y\le\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{4}\) và dấu ''='' xảy ra khi c = 4; a = 6; b = 8
6 tháng 5 2022
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\)
\(\le\dfrac{a}{2a\sqrt{bc}}+\dfrac{b}{2b\sqrt{ca}}+\dfrac{c}{2c\sqrt{ab}}\)
\(=\dfrac{a\sqrt{bc}}{2abc}+\dfrac{b\sqrt{ca}}{2abc}+\dfrac{c\sqrt{ab}}{2abc}\)
\(\le\dfrac{2a^2+b^2+c^2}{8abc}+\dfrac{2b^2+a^2+c^2}{8abc}+\dfrac{2c^2+b^2+a^2}{8abc}\)
\(=\dfrac{4\left(a^2+b^2+c^2\right)}{8abc}=\dfrac{1}{2}\)