cho a,b,c>0, dùng bđt cô si để chứng minh:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
K
Khách
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21 tháng 4 2018
Áp dụng BĐT Cô - si : x2 + y2 ≥ 2xy
=> \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\) ≥ \(2.\dfrac{a}{c}\) ( 1)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\) ≥ \(2.\dfrac{b}{a}\) ( 2)
\(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\) ≥ \(2.\dfrac{c}{b}\) ( 3)
Cộng từng vế của ( 1 , 3 , 3) , ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\) ≥ \(2.\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)
=> ĐPCM
6 tháng 4 2017
Áp dụng bđt Cauchy Schwarz dạng Engel ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\ge\left(a+b+c\right).\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\)
\(\ge\dfrac{9}{2}\left(đpcm\right)\)
17 tháng 3 2018
a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
21 tháng 6 2017
làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)
21 tháng 6 2017
ok thỏa thuận rồi tui làm nửa sau thui nhé :D
Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:
\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)
Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:
\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)
\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)
Can you continue
K
6 tháng 5 2018
a, b, c > 0
Áp dụng bất đẳng thức AM - GM (Cauchy):
\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\)
\(\dfrac{b}{c^2}+\dfrac{1}{b}\ge2\sqrt{\dfrac{b}{c^2}.\dfrac{1}{b}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\)
\(\dfrac{c}{a^2}+\dfrac{1}{c}\ge2\sqrt{\dfrac{c}{a^2}.\dfrac{1}{c}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\)
Vậy ta có :
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{b}+\dfrac{2}{c}+\dfrac{2}{a}\)
\(\Leftrightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) (đpcm)
HN
7 tháng 5 2018
Cách dùng hằng đẳng thức:
\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}-\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\)
\(=\left(\dfrac{a}{b^2}-\dfrac{2}{b}+\dfrac{1}{a}\right)+\left(\dfrac{b}{c^2}-\dfrac{2}{c}+\dfrac{1}{b}\right)+\left(\dfrac{c}{a^2}-\dfrac{2}{a}+\dfrac{1}{c}\right)\)
\(=\left(\dfrac{\sqrt{a}}{b}-\dfrac{1}{\sqrt{a}}\right)^2+\left(\dfrac{\sqrt{b}}{c}-\dfrac{1}{\sqrt{b}}\right)^2+\left(\dfrac{\sqrt{c}}{a}-\dfrac{1}{\sqrt{c}}\right)^2\ge0\)
17 tháng 2 2018
áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)
tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)
suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
suy ra dpcm
dau = xay ra khi a=b=c
QT
20 tháng 5 2018
\(\sum\dfrac{a^3}{a^2+b^2}=a+b+c-\dfrac{ab^2}{a^2+b^2}-\dfrac{bc^2}{b^2+c^2}-\dfrac{ca^2}{c^2+a^2}\ge a+b+c-\dfrac{b}{2}-\dfrac{c}{2}-\dfrac{a}{2}=\dfrac{a+b+c}{2}\) Dấu "=" xảy ra khi: \(a=b=c\)
Áp dụng BĐT Cauchy cho 2 số dương:
\(\left\{{}\begin{matrix}\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2}{b}.b}=2a\\\dfrac{b^2}{c}+c\ge2\sqrt{\dfrac{b^2}{c}.c}=2b\\\dfrac{c^2}{a}+a\ge2\sqrt{\dfrac{c^2}{a}.a}=2c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c\ge2a+2b+2c\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\left(đpcm\right)\)
Dấu "=" xay ra \(\Leftrightarrow a=b=c\)
Áp dụng BĐT cosi cho 3 số a,b,c dương:
\(\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2b}{b}}=2a\\ \dfrac{b^2}{c}+c\ge2\sqrt{\dfrac{b^2c}{c}}=2b\\ \dfrac{c^2}{a}+a\ge2\sqrt{\dfrac{c^2a}{a}}=2c\)
Cộng vế theo vế 3 BĐT trên
\(\Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c\ge2\left(a+b+c\right)\\ \Leftrightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
Dấu \("="\Leftrightarrow a=b=c\)