\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\) . Chứng minh bất đẳng thức với ∀a,b,c ≥0
Mọi người giúp em với ạ .
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 :
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) với a,b,c > 0
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
chứng minh bất đẳng thức:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\) với\(a\ge b\ge c>0\)
chứng minh bất đẳng thức
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\) 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
!!!
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For \(a\geq b\geq c>0\) we obtain:
\(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\)
M.n giúp em với ạ !!!
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
nhớ tìm trước khi hỏi Câu hỏi của Trần Huỳnh Cẩm Hân - Toán lớp 8 | Học trực tuyến
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 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 \(a\ge b\ge c\)>0
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}-\dfrac{a-b}{2\left(b+c\right)}+\dfrac{a-c}{2\left(b+c\right)}-\dfrac{a-c}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{2}\cdot\dfrac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{2}\left(\dfrac{1}{\left(b+c\right)\left(c+a\right)}-\dfrac{1}{\left(a+b\right)\left(b+c\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(a+c\right)\left(b+c\right)}\ge0\)(luôn đúng)
\(\Rightarrowđpcm\)
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
\(a^2+b^2+c^2\ge ab+bc+ca\)