Cho a, b, c >0 chứng minh rằng \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)≥\(\frac{3}{2}\)
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13 tháng 8 2017
3) Đặt b+c=x;c+a=y;a+b=z.
=>a=(y+z-x)/2 ; b=(x+z-y)/2 ; c=(x+y-z)/2
BĐT cần CM <=> \(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)
VT=\(\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1+\frac{x}{z}+\frac{y}{z}-1\right)\)
\(=\frac{1}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)-3\right]\)
\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)(Cauchy)
Dấu''='' tự giải ra nhá
13 tháng 8 2017
Bài 4
dễ chứng minh \(\left(a+b\right)^2\ge4ab;\left(b+c\right)^2\ge4bc;\left(a+c\right)^2\ge4ac\)
\(\Rightarrow\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2\ge64a^2b^2c^2\)
rồi khai căn ra \(\Rightarrow\)dpcm.
đấu " = " xảy ra \(\Leftrightarrow\)\(a=b=c\)
22 tháng 3 2019
Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
TN
31 tháng 7 2017
Áp dụng BĐT AM-GM ta có:
\(\frac{a}{1+b^2}=a-\frac{a^2b}{b^2+1}\ge a-\frac{a^2b}{2b}=a-\frac{ab}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{c^2+1}\ge b-\frac{bc}{2};\frac{c}{a^2+1}\ge c-\frac{ca}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge a+b+c-\frac{ab+bc+ca}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
31 tháng 7 2017
tc \(x^2+y^2\ge2xy\left(cauchy\right)\)
\(\frac{a}{1+b^2}=\frac{a+ab^2-ab^2}{1+b^2}=\frac{a\left(1+b^2\right)-ab}{1+b^2}=a-\frac{ab}{1+b^2}\ge a-\frac{ab}{2ab}\ge a-\frac{1}{2}\)(1)
tương tự \(\frac{b}{1+c^2}\ge b-\frac{1}{2}\)(2)
\(\frac{c}{1+a^2}\ge c-\frac{1}{2}\)(3)
từ (1)(2)(3)=> \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{3}{2}=3-\frac{3}{2}=\frac{3}{2}\left(a+b+c=3\right)\)
=> đpcm
CV
15 tháng 12 2016
áp dụng dbt cosi cho 2 số:\(\frac{a^3}{b^2}\)va b ta duoc :
\(\frac{a^3}{b^2}\)+a\(\ge\)2\(\sqrt{\frac{a^3}{b^2}.a}\)=2\(\frac{a^2}{b}\)
CMTT:\(\frac{b^3}{c^2}\)+b\(\ge\)2\(\frac{b^2}{c}\)
\(\frac{c^3}{a^2}\)+c\(\ge\)2\(\frac{c^2}{a}\)
\(\Rightarrow\)\(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{c^3}{a^2}\)+(a+b+c)\(\ge\)2(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\))
\(\Leftrightarrow\)\(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{c^3}{a^2}\)\(\ge\)2(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)) - (a+b+c) (1)
Ap dụng bdt cosi cho các số dương , ta được:
\(\frac{a^2}{b}\)+\(b\)\(\ge\)2\(\sqrt{\frac{a^2}{b}.b}\)=2a
CMTT: \(\frac{b^2}{c}\)+c\(\ge\)2b
\(\frac{c^2}{a}\)+a\(\ge\)2c
\(\Rightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)+(a+b+c) \(\ge\)2(a+b+c)
\(\Leftrightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\)\(\ge\)a+b+c
\(\Leftrightarrow\)\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\) _ (a + b + c ) \(\ge\)0
Do Đó:TỪ (1) ta co : \(\frac{a^3}{b^2}\)+\(\frac{b^3}{c^2}\)+\(\frac{b^3}{c^2}\)\(\ge\)(\(\frac{a^2}{b}\)+\(\frac{b^2}{c}\)+\(\frac{c^2}{a}\) )
T
29 tháng 5 2019
Xét hiệu hai vế:
BĐT \(\Leftrightarrow\left(\frac{a^3}{b^2}-\frac{a^2b}{b^2}\right)+\left(\frac{b^3}{c^2}-\frac{b^2c}{c^2}\right)+\left(\frac{c^3}{a^2}-\frac{c^2a}{a^2}\right)-\left(a+b+c-b-c-a\right)\ge0\)
\(\Leftrightarrow\left(\frac{a^3}{b^2}-\frac{a^2b}{b^2}\right)+\left(\frac{b^3}{c^2}-\frac{b^2c}{c^2}\right)+\left(\frac{c^3}{a^2}-\frac{c^2a}{a^2}\right)-\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]\ge0\)
\(\Leftrightarrow\left(\frac{a^2}{b^2}\left(a-b\right)-\left(a-b\right)\right)+\left(\frac{b^2}{c^2}\left(b-c\right)-\left(b-c\right)\right)+\left(\frac{c^2}{a^2}\left(c-a\right)-\left(c-a\right)\right)\ge0\)
\(\Leftrightarrow\frac{\left(a+b\right)\left(a-b\right)^2}{b^2}+\frac{\left(b+c\right)\left(b-c\right)^2}{c^2}+\frac{\left(c+a\right)\left(c-a\right)^2}{a^2}\ge0\)
BĐT này đúng với mọi a,b,c > 0 nên ta có Q.E.D
Dấu "=" xảy ra khi a =b =c
P/s: Toán 7 gì mà khó thế nhỉ??Mình cũng không chắc đâu nha!
2 tháng 7 2019
Áp dụng Svac
\(\Sigma\frac{a^3}{b+c}=\Sigma\frac{a^4}{ab+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\left(a^2+b^2+c^2\right)\)
"=" tại a=b=c
4 tháng 7 2019
E thử làm cách khác ạ:))
Không mất tính tổng quát,giả sử \(a\ge b\ge c\)
\(\Rightarrow\hept{\begin{cases}a^2\ge b^2\ge c^2\\\frac{a}{b+c}\ge\frac{b}{a+c}\ge\frac{c}{a+b}\end{cases}}\)
Áp dụng BĐT Trebysev ta có:
\(a^2\cdot\frac{a}{b+c}+b^2\cdot\frac{b}{a+c}+c^2\cdot\frac{c}{a+b}\ge\frac{a^2+b^2+c^2}{3}\cdot\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{3}{2}\left(nesbitt\right)\)
\(=\frac{a^2+b^2+c^2}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
8 tháng 5 2018
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\)
=> \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Bài làm
Đặt x = a + b , y = b + c , z = c + a
Thì \(a=\frac{x+z-y}{2};b=\frac{x+y-z}{2};c=\frac{y+z-x}{2}\)
Ta có: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{\frac{x+z-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}+\frac{\frac{y+z-x}{2}}{x}\)
\(\Leftrightarrow\frac{x+z-y}{2}.\frac{1}{y}+\frac{x+y-z}{2}.\frac{1}{z}+\frac{y+z-x}{2}.\frac{1}{x}\)
\(\Leftrightarrow\frac{x+z-y}{2y}+\frac{x+y-z}{2z}+\frac{y+z-x}{2x}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{z}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}-3\right)\)
\(\Leftrightarrow-3.\frac{1}{2}+\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)\)
\(\Leftrightarrow-\frac{3}{2}+\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)\ge\frac{1}{2}.6-\frac{3}{2}=\frac{3}{2}\) ( đpcm )
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