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Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
\(\dfrac{1}{a+3b}+\dfrac{1}{a+b+2c}\ge\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)
Tương tự: \(\dfrac{1}{b+3c}+\dfrac{1}{b+c+2a}\ge\dfrac{2}{a+b+2c}\)
\(\dfrac{1}{c+3a}+\dfrac{1}{a+c+2b}\ge\dfrac{2}{2a+b+c}\)
Cộng vế với vế và rút gọn:
\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)
\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)
*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)
Nhận xét: Với x,y > 0 ta có:
\(4xy\le\left(x+y\right)^2\)
<=> \(\dfrac{1}{x+y}\le\dfrac{x+y}{4xy}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Xảy ra khi x = y
Áp dụng và bài ta có:
\(\dfrac{1}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\)
Tương tự: \(\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\);
\(\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\)
Cộng 3 vế bđt có:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{3}{4}\)
Hì hì, thật ra thì mình không biết giúp thằng bạn mình như thế nào nên đành tự đăng câu hỏi vậy :))
Giả sử c là số ở giửa a và b. khi đó \(\left(b-c\right)\left(c-a\right)\ge0\)
Ta chứng minh :
\(VT\le c\left(\dfrac{b^2}{2b^2+a^2+c^2}+\dfrac{a^2}{2a^2+b^2+c^2}\right)+\dfrac{abc}{a^2+b^2+2c^2}\)(*)
\(\Leftrightarrow\dfrac{\left(c-a\right)\left(b-c\right)\left(b^2+c^2-bc+a^2\right)}{\left(a^2+c^2+2b^2\right)\left(b^2+a^2+2c^2\right)}\ge0\) (Đúng)
Áp dụng BĐT AM-GM:
\(VT\le\dfrac{c}{4}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}+\dfrac{a^2}{a^2+c^2}\right)+\dfrac{abc}{2ac+2bc}\)
\(\le\dfrac{c}{4}\left(1+\dfrac{b^2}{2bc}+\dfrac{a^2}{2ac}\right)+\dfrac{\dfrac{\left(a+b\right)^2}{4}}{2\left(a+b\right)}=\dfrac{c}{4}+\dfrac{a+b}{8}+\dfrac{a+b}{8}\)
\(=\dfrac{a+b+c}{4}\)( \(ĐpcM\))
Dấu = xảy ra khi a=b=c
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm
Câu a)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq \frac{9}{a+2b}\) (1)
\(\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\geq \frac{9}{b+2c}\)(2)
\(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\geq \frac{9}{c+2a}\) (3)
Lấy \((1)+2.(2)+3.(3)\) ta có:
\(\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{2}{b}+\frac{2}{c}+\frac{2}{c}+\frac{3}{c}+\frac{3}{a}+\frac{3}{a}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)
\(\Leftrightarrow \frac{7}{a}+\frac{4}{b}+\frac{7}{c}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
Câu b)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a}+\frac{4}{b}\geq \frac{(1+2)^2}{a+b}=\frac{9}{a+b}\)
\(\Rightarrow \frac{1}{3a}+\frac{4}{3b}\geq \frac{3}{a+b}(1)\)
\(\frac{1}{3b}+\frac{1}{2c}+\frac{1}{2c}\geq \frac{9}{3b+4c}\)
\(\Rightarrow \frac{2}{3b}+\frac{2}{c}\geq \frac{18}{3b+4c}\) (2)
\(\frac{1}{c}+\frac{1}{3a}+\frac{1}{3a}\geq \frac{9}{c+6a}\) (3)
Từ (1); (2); (3) cộng theo vế:
\(\Rightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{3}{a+b}+\frac{18}{3b+4c}+\frac{9}{c+6a}\)
(đpcm)
Dấu bằng xảy ra khi \(a=\frac{b}{2}=\frac{c}{3}\)
Câu c)
BĐT cần chứng minh tương đương với:
\(\frac{b+c+a}{a}+\frac{2a+c}{b}+\frac{4(a+b)}{a+c}\geq 10\) (*)
Áp dụng BĐT AM-GM:
\(\text{VT}=\frac{b}{a}+\frac{c+a}{2a}+\frac{c+a}{2a}+\frac{a}{b}+\frac{a+c}{2b}+\frac{a+c}{2b}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}\)
\(\geq 10\sqrt[10]{\frac{ba(c+a)^4(a+b)^4}{16a^3b^3(a+c)^4}}=10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\)
Theo AM-GM: \((a+b)^2\geq 4ab\Rightarrow (a+b)^4\geq 16a^2b^2\)
\(\Rightarrow \text{VT}\geq 10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\geq 10\)
Vậy (*) được cm. Ta có đpcm. Dấu bằng xảy ra khi a=b=c
Theeo BĐT AM-GM ta có:
\(\sum\dfrac{a^3b}{a^4+a^2b^2+b^4}\le\sum\dfrac{a^3b}{2a^3b+b^4}=\sum\dfrac{a^3}{2a^3+b^3}\)
Ta cần chứng minh \(\sum\dfrac{a^3}{2a^3+b^3}\le1\)
hay \(\sum\dfrac{a^3}{a^3+2c^3}\ge1\)
Áp dụng BĐT Cauchy - Schwarz có:
\(\sum\dfrac{a^3}{2c^3+a^3}\ge\dfrac{\left(\sum a^3\right)^2}{\sum a^6+2\sum a^3b^3}=1\)
Đẳng thức xảy ra khi a = b = c