Cho: \(\left\{{}\begin{matrix}a,b,c>0\\a^2+b^2+c^2=abc\end{matrix}\right.\)
Tìm Max:
\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ac}+\dfrac{c}{c^2+ab}\)
Cho \(\left\{{}\begin{matrix}a;b;c>0\\ab+bc+ac=1\end{matrix}\right.\) Tìm max của \(P=\dfrac{1-a^2}{1+a^2}+\dfrac{1-b^2}{1+b^2}+\dfrac{1-c^2}{1+c^2}\)
Cái c là \(\dfrac{2}{\sqrt{1+c^2}}\) ạ
\(P=\dfrac{2-\left(1+a^2\right)}{1+a^2}+\dfrac{2-\left(1+b^2\right)}{1+b^2}+\dfrac{2}{\sqrt{1+c^2}}\)
\(P=2\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{\sqrt{1+c^2}}\right)-2\)
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 \(\left\{{}\begin{matrix}x=y=\dfrac{\pi}{8}\\z=\dfrac{\pi}{4}\end{matrix}\right.\) \(\Rightarrow\left(a;b;c\right)=\left(\sqrt{2}-1;\sqrt{2}-1;1\right)\)
cho:\(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\)
CMR: \(A=\Sigma\dfrac{1}{5a^2+ab+bc}\ge\dfrac{3}{7}\)
cho \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le\dfrac{3}{2}\end{matrix}\right.\)
tìm \(MinS=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
Lâu rồi không lên Hoc24
Áp dụng bất đẳng thức Minkowski, Schwarz và AM - GM ta có:
\(S\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}=\sqrt{\left[\left(a+b+c\right)^2+\dfrac{81}{16\left(a+b+c\right)^2}\right]+\dfrac{81.15}{16\left(a+b+c\right)^2}}\ge\sqrt{\dfrac{9}{2}+\dfrac{135}{4}}=\sqrt{\dfrac{153}{4}}=\dfrac{3\sqrt{17}}{2}\).
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??
+) Giải hệ pt: \(\left\{{}\begin{matrix}4\sqrt{x^2+4y-5}=y^2-x+10\\x^3+\left(1-y\right)x^2=\left(x+4\right)y\end{matrix}\right.\)
+) Cho a,b,c>0 và a+b+c=2017
CM: \(\dfrac{2017a-a^2}{bc}+\dfrac{2017b-b^2}{ca}+\dfrac{2017c-c^2}{ab}\ge\sqrt{2}\left(\Sigma\sqrt{\dfrac{2017-a}{a}}\right)\)
+) 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)
Anh Trần Tuấn Hoàng giỏi BĐT quá nhỉ
cho a,b,c dương khác nhau đôi một thỏa \(\left\{{}\begin{matrix}ab+bc=2c^2\\2a\le c\end{matrix}\right.\)
tìm Max \(\dfrac{a}{a-b}+\dfrac{b}{b-c}+\dfrac{c}{c-a}\)
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 \(x=\dfrac{4}{3}\)
CMR: \(\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ca}\ge670\) với \(\left\{{}\begin{matrix}a+b+c\le3\\a,b,c>0\end{matrix}\right.\)
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
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{2009}{ab+bc+ac}\ge669+1=670\left(dpcm\right)\)
Cho a, b, c thỏa mãn \(\left\{{}\begin{matrix}a\ge3\\b\ge4\\c\ge2\end{matrix}\right.\)
Tìm Max \(y=\dfrac{ab\sqrt{c-2}+bc\sqrt{a-3}+ac\sqrt{b-4}}{abc}\)
\(\Leftrightarrow y=\dfrac{\sqrt{c-2}}{c}+\dfrac{\sqrt{a-3}}{a}+\dfrac{\sqrt{b-4}}{b}\)
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
Cho số thực a < 0 và hai tập hợp A = (-∞; 9a), B = (\(\dfrac{4}{a}\); +∞). Tìm a để A\(\cap\)B ≠ ∅
A. \(\left[{}\begin{matrix}a\ge3\\a< -4\end{matrix}\right.\)
B. \(\left[{}\begin{matrix}a\ge\dfrac{5}{2}\\a< -\dfrac{1}{3}\end{matrix}\right.\)
C. \(\left[{}\begin{matrix}a< \dfrac{5}{2}\\a\ge-\dfrac{1}{3}\end{matrix}\right.\)
D. -\(\dfrac{1}{3}\)≤ a ≤ \(\dfrac{5}{2}\)
1. Cho a,b,c t/m: \(\left\{{}\begin{matrix}a\ge\dfrac{4}{3}\\b\ge\dfrac{4}{3}\\c\ge\dfrac{4}{3}\end{matrix}\right.\) và \(a+b+c=6\)
\(CMR:\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\ge\dfrac{6}{5}\)
2. Cho x,y >0 t/m: \(2x+3y-13\ge0\)
Tìm min \(P=x^2+3x+\dfrac{4}{x}+y^2+\dfrac{9}{y}\)
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 \("="\Leftrightarrow a=b=c=2\)