chứng minh bất đẳng thức
A=(a+b)(\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\))\(\ge\) 4
help me!!TT
b) B=\(\dfrac{a+b}{c}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{c+a}{b}\)\(\ge\)6 ( a,b,c > 0)
Chứng minh bất đẳng thức sau:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(a,b,c>0\right)\)
Áp dụng BĐT cosi:
\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)
Dấu \("="\Leftrightarrow a=b=c\)
chứng minh bất đẳng thức
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\)với a ≥ b ≥ c > 0
Ta có: BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{2a-\left(a+b\right)}{2\left(a+b\right)}+\dfrac{2b-\left(b+c\right)}{2\left(b+c\right)}+\dfrac{2c-\left(c+a\right)}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{\left(b+c\right)\left(c+a\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\) (đúng)
Vậy BĐT luôn đúng với \(a\ge b\ge c>0\)
Chứng minh bất đẳng thức : \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\) vs \(a\ge b\ge c>0\)
Ta có: \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{a}{2b}+\dfrac{b}{2c}+\dfrac{c}{2a}=\dfrac{1}{2}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\ge\dfrac{1}{2}.3=\dfrac{3}{2}\) ( BĐT AM - GM )
Dấu " = " khi a = b = c
\(\Rightarrowđpcm\)
BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{2}\cdot\dfrac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{\left(b+c\right)\left(c+a\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{\left(c-a\right)\left(c+a\right)+\left(a-c\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(luôn đúng)
\(\Rightarrowđpcm\)
chứng minh bất đẳng thức:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\) với\(a\ge b\ge c>0\)
Sử dụng bất đẳng thức cô-si. Chứng minh bất đẳng thức \(\dfrac{a}{bc}+\dfrac{c}{ba}+\dfrac{b}{ac}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Coi như a, b, c là số dương
Áp dụng BĐT Cô-si ta có:
\(\dfrac{a}{bc}+\dfrac{c}{ba}\ge2\sqrt{\dfrac{a}{bc}.\dfrac{c}{ba}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\left(1\right)\)
Dấu "=" xảy ra ...
\(\dfrac{a}{bc}+\dfrac{b}{ac}\ge2\sqrt{\dfrac{a}{bc}.\dfrac{b}{ac}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\left(2\right)\)
Dấu "=" xảy ra ...
\(\dfrac{c}{ba}+\dfrac{b}{ac}\ge2\sqrt{\dfrac{c}{ba}+\dfrac{b}{ac}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\left(3\right)\)
Dấu "=" xảy ra ...
Từ (1), (2), (3) ta có:
\(\dfrac{a}{bc}+\dfrac{c}{ba}+\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ba}+\dfrac{b}{ac}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\\ \Rightarrow2\left(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ba}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ \Rightarrow\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ba}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu "=" xảy ra ...
Vậy ...
a, b, c có phải là số dương không bạn, nếu không thì làm sao dùng BĐT Cô-si được
Cho a ≥ b ≥ c >0.
Chứng minh bất đẳng thức: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\) ≤ \(\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{b}\)
chứng minh bất đẳng thức
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\) với \(a\ge b\ge c>0\)
Tham khảo ở đây có đủ các cách cho bạn chọn lựa
Từ "Siêu tốc thần sầu" đến "tập thể dục" tha hồ luyện
!!!
https://hoc24.vn/hoi-dap/question/196314.html
For \(a\geq b\geq c>0\) we obtain:
\(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\)
Áp dụng bất đẳng thức cosi chứng minh
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với a,b \(\ge\)0
\(\left(a+b\right).\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\) 4 với a,b > 0
\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\) 9 với a,b,c > 0
\(a^2+b^2+c^2\ge ab+bc+ca\)
Chứng minh bất đẳng thức: \(\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (với a, b, c > 0)
BĐT cần chứng minh tương đương :
\(\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\ge\dfrac{ab+bc+ac}{abc}\)
\(\Leftrightarrow\dfrac{a^8+b^8+c^8}{a^2b^2c^2}\ge ab+bc+ac\)
\(\Leftrightarrow\dfrac{a^6}{b^2c^2}+\dfrac{b^6}{a^2c^2}+\dfrac{c^6}{a^2b^2}\ge ab+bc+ac\)
Do \(a^2+b^2+c^2\ge ab+bc+ac\)
Ta phải cm
\(\dfrac{a^6}{b^2c^2}+\dfrac{b^6}{a^2c^2}+\dfrac{c^6}{a^2b^2}\ge a^2+b^2+c^2\)(1)
Đặt : \(\left(a^2;b^2;c^2\right)=\left(x;y;z\right)\)
\(\Rightarrow\left(1\right)\Leftrightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}\ge x+y+z\)
Áp dụng C.B.S
\(\Rightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}=\dfrac{x^4}{xyz}+\dfrac{y^4}{xyz}+\dfrac{z^4}{xyz}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3xyz}\)
Theo Bunhiacopxki: \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\)\(\Rightarrow\left(x^2+y^2+z^2\right)^2\ge\dfrac{\left(x+y+z\right)^4}{9}\)
Theo Cauchy : \(\Rightarrow3xyz\le\dfrac{\left(x+y+z\right)^3}{9}\)
\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3xyz}\ge\dfrac{\dfrac{\left(x+y+z\right)^4}{9}}{\dfrac{\left(x+y+z\right)^3}{9}}=x+y+z\)
\(\Rightarrow\)\(\Rightarrow\dfrac{x^3}{yz}+\dfrac{y^3}{xz}+\dfrac{z^3}{xy}\ge x+y+z\)
=> đpcm